How much work is required to push a chair across the floor with a force of F = 8 lb ⁡ F=8\operatorname{lb} from x = 6 ft ⁡ x=6\operatorname{ft} to x = 9 ft ⁡ x=9\operatorname{ft} along the x x -axis?

24 ft ⁡ ⋅ lb ⁡ s 24\operatorname{ft}\cdot\operatorname{lb}s

120 ft ⁡ ⋅ lb ⁡ s 120\operatorname{ft}\cdot\operatorname{lb}s

72 ft ⁡ ⋅ lb ⁡ s 72\operatorname{ft}\cdot\operatorname{lb}s

48 ft ⁡ ⋅ lb ⁡ s 48\operatorname{ft}\cdot\operatorname{lb}s

How much work is done by a person lifting a 10 lb ⁡ 10\operatorname{lb} bucket 4 ft ⁡ 4\operatorname{ft} off the ground?

40 ft ⁡ ⋅ lb ⁡ s 40\operatorname{ft}\cdot\operatorname{lb}s

100 ft ⁡ ⋅ lb ⁡ s 100\operatorname{ft}\cdot\operatorname{lb}s

0 ft ⁡ ⋅ lb ⁡ s 0\operatorname{ft}\cdot\operatorname{lb}s

400 ft ⁡ ⋅ lb ⁡ s 400\operatorname{ft}\cdot\operatorname{lb}s

A 120 m ⁡ 120\operatorname{m} chain hangs freely from the side of a building. The chain weighs 15 kg ⁡ 15\operatorname{kg} / m m . How much work is done to pull 80 m ⁡ 80\operatorname{m} of the chain to the top of the building?

940800 J ⁡ 940800\operatorname{J}

9600 J ⁡ 9600\operatorname{J}

4800 J ⁡ 4800\operatorname{J}

470400 J ⁡ 470400\operatorname{J}

Find the work done by fully winding up a cable of length 30 ft ⁡ 30\operatorname{ft} and weight-density 2 lb ⁡ 2\operatorname{lb} / ft ⁡ \operatorname{ft} .

9000 J ⁡ 9000\operatorname{J}

A 40 m ⁡ 40\operatorname{m} rope hangs freely over a ledge. The density of the rope is 12 kg ⁡ 12\operatorname{kg} / m m . If a 5 kg ⁡ 5\operatorname{kg} bucket is attached to the end of the rope, how much work is done to pull the rope and the bucket to the ledge?

96040 J ⁡ 96040\operatorname{J}

94080 J ⁡ 94080\operatorname{J}

1960 J ⁡ 1960\operatorname{J}

92120 J ⁡ 92120\operatorname{J}

A 60 ft ⁡ 60\operatorname{ft} cable is attached to a cylinder that is attached to a winch. If the cable weighs 300 lbs, how much work is needed to wind 20 ft ⁡ 20\operatorname{ft} of the cable onto the cylinder using the winch? Hint: Divide cable weight by cable length to get density.

5000 J ⁡ 5000\operatorname{J}

3000 J ⁡ 3000\operatorname{J}

17900 J ⁡ 17900\operatorname{J}

1700 J ⁡ 1700\operatorname{J}

A swimming pool has the shape of a rectangular prism with abase that measures 30 ft ⁡ \operatorname{ft} by 20 ft ⁡ \operatorname{ft} and is 5 ft ⁡ \operatorname{ft} deep. The top of the pool is 1 ft ⁡ \operatorname{ft} above the surface of the water. How much work is required to pump all the water out? Assume the density of water is 62.4 lb ⁡ \operatorname{lb} / ft ⁡ 3 \operatorname{ft}^3 .

449280 ft ⁡ ∙ lb ⁡ s \operatorname{ft}\bullet\operatorname{lb}s

42120 ft ⁡ ∙ lb ⁡ s \operatorname{ft}\bullet\operatorname{lb}s ft ∙ lb s

79560 ft ⁡ ∙ lb ⁡ s \operatorname{ft}\bullet\operatorname{lb}s ft ∙ lb s

280800 ft ⁡ ∙ lb ⁡ s \operatorname{ft}\bullet\operatorname{lb}s ft ∙ lb s

10. Physics Applications of Integrals

Work

10. Physics Applications of Integrals

Work: Videos & Practice Problems

Learn Concepts Practice Worksheet

Topic summary

Calculus can be applied to real-world scenarios such as work problems, where work is done when a force is applied over a distance. For constant forces, work is calculated using the equation $W = F d$ . For variable forces, integration is used to find the area under the force curve. Hooke's Law describes work done on springs, while lifting and pumping problems involve integrating force functions over distances. Understanding these concepts enhances problem-solving skills in physics and engineering.

concept

Introduction To Work

Video duration:

Video transcript

In recent videos, we've been spending a lot of time talking about how to apply calculus to the real world. This happened when we dealt with things like acceleration, velocity, or position. Now it turns out there are quite a few other scenarios in this course where we can apply calculus to the real world. And in this video, we're going to be taking a look specifically at work problems. Work is going to be done on an object when a force is applied to that object over a certain distance.

When this happens, you have work being done, and I'm going to show you the strategies we use to calculate work and solve these problems. So let's get right into things. Now let's say, for example, we had a box, and we were applying a force to this box. If we applied that force to move the box a certain distance, we could say that we did work on this box. Now if your force is given in newtons and your distance is given in meters, then the work that you calculate will be measured in joules.

Now, likewise, if you have pounds as the force that you're dealing with and feet as the distance that you're measuring, then your work is going to be measured in a unit known as foot-pounds. So this is just something to be aware of when you solve these types of problems. Let's say we have this situation where we're applying a force to a box, and we're dealing with an example that says find the work done on the object by the force given in newtons from x=1 to x=4 m2. Notice we have a clear force that we're applying, and we have a certain distance that we're applying the force, meaning that we are doing work on the object. If you want to calculate work and find a number for this, work is going to be the area underneath the force curve.

For example, if we were dealing with a constant force of 22 newtons. This would be an example of a constant force because we can see this force is continually applied and doesn't change. We're asked to find the work done from x=1 to x=4 meters. If we're trying to find the area under this curve as mentioned before, then we're just finding this area from x=1 all the way over to x=4. This is the area that we're trying to calculate right here.

We know some strategies when it comes to finding the area under a function curve. One of the strategies we've learned is to use an integral to figure this out. But you may notice something about this specific example. This is a familiar shape. We have seen this shape before. It looks like a rectangle. So we can actually just use the area of a rectangle to solve this problem. I know the area of a rectangle is just going to be base times height, where this is the base and that's the height. Clearly, in this problem, my height is just going to be the force function that's constant. So in this case, I'm just going to have force multiplied by the base here, which is just going to be the distance that we travel on the x-axis.

Force times distance is the equation that we can use when it comes to dealing with these work problems where we have a constant force. To solve this specific example, I'm going to take our force of 22 newtons and multiply it by our distance. I see our distance goes from one all the way over to four, and the difference between four and one is going to be three. We need to travel three units to get to four from one. So that means we'll have 22 times three, which turns out to be 66.

The unit for this problem is going to be joules since the force we're dealing with is in newtons and the distance we're dealing with is in meters. That right there is the solution to this problem. Now this all works when you are dealing with constant forces. But what if instead we had a force that wasn't constant? What would that really mean?

A non-constant force would simply mean that we're not just continually applying the same force over that distance. Perhaps we're applying a certain amount of force here, but maybe we're applying an even greater force at this point. If this were the case, what we would need to do is recognize how that force is changing over the distance that we apply the force to that object. In that case, we could see a function like this. Notice here, it's clear to us that we are applying more and more force as we travel this distance, and we're still trying to find the work done from x=1 to x=4 meters.

I can use the same strategy and recognize that the work is going to be the area underneath the curve of the force function. This would be the area right here. But the unfortunate thing is with this example is now there's not really a familiar shape that forms. It kind of looks like a triangle, but that's not exact. A rectangle is not going to fit the bill here either.

So what we need to do is use an integral to solve this problem. If I'm dealing with a work situation where I don't have a constant force, what we need to do is integrate our work from whatever our start distance is to our finish distance, and then that's going to be the force function that we're integrating. If I'm looking to solve this problem down here, well, I'm going to be finding the work from one to four meters. So that's going to go from one to four. That's my integral.

The force function I can see is x2. So all we're doing is finding this integral right here. Now I know how to do this, I know the general process. I know that first I need to find the antiderivative of x2, which using the power rule is going to be x33. Now from here I would need to bind this function from one all the way to four.

Now this process we should know how to do. We first take our high bound, plug it in, then we take our low bound and plug it in, and then we subtract the two results. If you do that, you should get 21 as your answer, and that is going to be in joules. So 21 joules is going to be the solution to this problem, and that's how we can solve problems when dealing with a variable force. So notice that this process for solving work is always just finding the area underneath the function curve.

And keep in mind here that we don't actually even need to use this equation. If we wanted to, we could have integrated both these equations since we know that strategy always works for finding the area under the curve. However, this equation just serves as a nice little shortcut here when we know we're dealing with a constant force when we see that that force isn't going to vary at all in the problem. And since this will happen in a lot of different scenarios, it's really good to know this equation as well. But the main thing to remember is that when it comes to calculating work, it's always going to be the area underneath the curve of the force function, assuming that these graphs have force on the vertical axis and distance on the horizontal axis.

So I hope you found this video helpful, and let's try getting some more practice and solving some more problems. See you in the next one.

Problem

How much work is required to push a chair across the floor with a force of $F equals 8 lb$ from $x equals 6 ft$ to $x equals 9 ft$ along the $x$ -axis?

$120 ft times lb of s$

$72 ft times lb of s$

$24 ft times lb of s$

$48 ft times lb of s$

Problem

How much work is done by a person lifting a $10 lb$ bucket $4 ft$ off the ground?

$100 ft times lb of s$

$0 ft times lb of s$

$40 ft times lb of s$

$400 ft times lb of s$

Problem

How much work is required to move an object with a force of $F of x equals 4 x cubed N$ acting along the $x$ -axis from $x equals 0 m$ to $x equals 2 m$ ?

$48 J$

$16 J$

$64 J$

$4 J$

Problem

Compute the work done by a force of $F = \frac{3}{x^{2}} N$ from $x equals 2 m$ to $x equals 6 m$ .

$1 J$

$negative 0.12 J$

$0.36 J$

$0.67 J$

concept

Work Done On A Spring (Hooke's Law)

Video duration:

Video transcript

So in recent videos, we've been introduced to the concept of work when it comes to solving various problems in calculus. And you may recall that the work equation looks like this, where we have to integrate our force function over a certain distance. That's what we do when it comes to work is we find the area under the curve of that force function, which we can do with this integral. Now it turns out a specific type of work problem you're going to see in this course is dealing with springs and finding the work done on that spring, which will also at times be referred to as Hooke's Law in your textbook. So that's what we're going to take a look at in this video.

So without further ado, let's jump right in. Now we can recall how to find the work acting on an object from point a to point b with this integral that was described before. Now it turns out that we can use this strategy when it comes to finding the work done specifically on a spring. So let's say we're trying to find the work when it comes to either stretching or compressing a spring. Well, if we do this, the work that we're going to do depends on the force that we apply to the spring, and the force we apply to the spring depends on how stiff that spring is.

The stiffness of the spring is measured by the spring constant k. Now this k right here can be measured in different units; it could be measured in newtons per meter or in pounds per foot, depending on what units you use. But notice how we always have a force over a distance because what we're trying to figure out is how much force we need to apply over the distance we stretch that spring. That tells us how stiff that spring is.

Now another thing to be familiar with is this point right here, which is known as the equilibrium point. The equilibrium point tells us where the spring is going to stretch. Right? We can't just set our point over here because we can see that the spring would actually be very compressed if we were at that point right there. So instead, we have to be where the spring naturally rests to be at equilibrium.

Now if we stretch the spring past equilibrium or compress it, we could say we're doing work on the spring since we have to apply a force over however much distance we stretch or compress the spring. So let's actually try applying this to an example to see if we could solve a situation where we're dealing with the spring. So in this example, we're told, given a spring constant k equals three newtons per meter, notice we have a force over a distance, that's our stiffness right there, how much work is done to stretch the spring from four meters to six meters? So we're trying to calculate work in this problem given this information on a spring. Now, first off, we should really be familiar with what equations we can use to solve this.

I mean, we know that the work equation is just going to be integrating this force function here, but what exactly would that force function be? Well, to find the force function when it comes to dealing with the spring, all you want to do is take your spring constant k and multiply it by the distance that it's stretched, and that's going to be a function with respect to this distance right here. And I think that makes sense because this would mean that if you want to stretch the spring super far, like, say, a hundred meters past equilibrium point, then you would need to apply a really great force to do that, a really strong force. So it makes sense that these would be the two variables that it's dependent on. Now when it comes to solving the work done on a spring, our first step should be to take our k value and plug it into this force function that we just discussed.

Now it's also possible that you'll see scenarios where this spring constant is not given. And if k is not given to you, then what you can do is take the force that you apply to the spring and divide it by the distance. But since we can see that this k value is already given to us in the problem, we don't need to do that. So all we need to do is plug this value of three in for k. So doing this, we're going to get that the force function is equal to three, because that's our spring constant up here, multiplied by x.

So that's our function right there. Now our next step is going to be determining the bounds of this function. So the bounds are going to be the a and b. Now a is the starting distance from the equilibrium point, and then b is the end distance from equilibrium. So if we go ahead and want to set up these bounds, well, I can see that the equilibrium point, we're going to be stretching from four meters all the way to six meters.

So four meters is clearly going to be the closest distance to our equilibrium, where six meters is going to be the farthest distance. So setting up my integral here, that means that we're going to go from four all the way up to six, and those are going to be the bounds. Now next, remember, we're doing the work equation right here, so I need to plug in my force function, which I figured out was 3x. So this is the setup to the integral. Now our last step is simply going to be solving this integral to find the work that we've done on this spring.

So I can go ahead and use the rules that I know about for integration. I can go ahead and use the power rule for three x to get three x squared over two. Then I'm going to bound this between four and six. Now first off, take up my high bound and plug it in for x. Then I'll take my low bound and plug it in for x and subtract the results.

So doing this, I'm going to get three times six squared, since I can see we have x squared on top, divided by two. Then we'll have minus three times four squared all divided by two. And if you go ahead and run these numbers right here, you can put those onto a calculator or do this by hand, you should get that your result is 30 joules. So 30 joules is going to be the work done to stretch the spring from four meters to six meters, and that's the solution to the problem. So this is the general process that you can take when it comes to solving these types of spring problems where you need to stretch or compress a spring.

So the main thing to remember is you first need to figure out your force function, and then once you do that, you just use the work equation to integrate this, and that will give you your result. So hope you found this video helpful, and let's move on and get some more practice.

Problem

A spring requires $12 J$ of work to stretch the spring from $1.1 m$ to $1.4 m$ past its equilibrium. What is the spring constant?

$0.375 N$ / $m$

$4.5 N$ / $m$

$12 N$ / $m$

$32 N$ / $m$

Problem

A spring requires a force of $8 N$ to stretch the spring to $5 cm$ past its equilibrium point. How much work could it take to stretch the spring from $2 cm$ to $9 cm$ past equilibrium?

$0.616 J$

$11.2 J$

$160 J$

$1.23 J$

Problem

Suppose a force of $10 N$ is required to stretch a spring $0.5 m$ from its equilibrium position. How much work is required to compress the spring $0.2 m$ from its equilibrium position?

$20 J$

$0.4 J$

$0.8 J$

$3 J$

example

Work Done On A Spring (Hooke's Law) Example 1

Video duration:

Video transcript

Let's try solving a more complicated example like this one. Here we're told a spring has a natural length of eight inches, and it takes 24 pounds of force to stretch the spring 12 inches. Our first part of the problem here asks us to find the spring constant k. Now recall that we can use Hooke's law, which says that force is equal to the spring constant times x. Now I can see that we're given the force in this problem that we can plug into this equation, but we need to figure out the distance here.

Now I'm going to go ahead and draw a sketch of what's happening here so you can kinda get an idea. So what we have is this spring, which is naturally going to be resting at eight inches. That's the natural length of the spring. So we say that eight inches is right here, meaning this would be our equilibrium point. Now what we're going to do is take this spring and we're going to stretch it to a distance of 12 inches so this would be a distance of 12 inches and our equilibrium would stay at the same point.

So notice that we started at eight inches and then we stretched it to 12 inches. Now I can see that we clearly needed to go four extra inches of length here, because 12 minus eight, that would be this four inch length right there. So four inches is how far we've stretched this spring. So we can say that our x here is going to be four inches. I can go ahead and clear this sketch right here because this now gives me a way to go about solving this problem where I need to find the spring constant. You can see my force here is 24 pounds, and then I can see we have our spring constant, which is unknown, and our distance x is going to be four inches.

Now a common mistake that's really easy to make at this point is to just go ahead and divide four on both sides and then solve from there. But there's something that's going to happen if we do that. Notice that the units for our distance are in inches, and we have 24 pounds. Typically, we want pounds and feet when we're dealing with these types of problems, because our work is going to be measured in foot-pounds. Well, we have four inches.

And what do I know about inches and feet? Well, the thing that I know about feet is that there are 12 inches in a foot. So one foot is equivalent to 12 inches. So what that means is I can go ahead and cancel these inches right here, giving me four over 12, which reduces to one third of a foot. So we can say that four inches is one third of a foot.

So that means that 24 is going to equal k times one third. Now from here, you can solve for this k by multiplying both sides of this equation by three. That's going to get the threes to cancel there, giving us that our spring constant k is equal to three times 24, which is 72, and that's going to be pounds per foot. So that's our spring constant k. As you can see, we now have proper units of pounds per foot, so we have solved part a of this problem.

You want to pay close attention to the units here, because that's what's going to allow you to solve problems when you see them. It's best to have pounds and feet that you're dealing with, and that's also best to or it's best to have newtons and meters depending on what your units are. So they can see we were given inches and pounds to be using this system of measurement. I know that I wanted pounds and feet for my measurement. So that's part a.

Now part b asks us how much work it would take to stretch the spring 12 inches to 16 inches. Now recall there's an equation we can use when calculating work. Work is going to be equal to the integral from a to b of our force function integrated with respect to x. Now the main thing is going to be figuring out these bounds right here, because I know my force function based on Hooke's law is just going to be the spring constant times x. We already calculated the spring constant to be 72, so it's going to be 72x, meaning that we're going to have the integral of 72x integrated with respect to x for our work. Now what exactly would the bounds be?

Well, to find the bounds, notice we go from 12 inches to 16 inches. But I can't just plug these in directly for my integral, because notice we're given inches. So how exactly do I solve this? Well, 12 inches, that's just a foot. So I can say we're starting at one foot, or that's going to be the lower bound or integral, I should say.

And then our top bound is going to be 16 inches. And what is 16 inches? Well, if 16 inches is this and we know that we can have one foot, which is 12 inches, well what our fraction is going to end up being is 16 over 12. And 16 over 12 reduces to four thirds. So we can say that 16 inches is equivalent to four thirds of a foot.

So this is how we integrate our function. Now to do this, I can pull the 72 outside the integral, and I know that the antiderivative of x is just going to be x² over two. So you're going to have x² over two integrated from one all the way up to four thirds. Now at this point, I need to plug in the bounds. You can start with my high bound of four thirds and plug that in.

So we're going to have 72. That's going to be multiplied by, in this case, plugging in four thirds. We're going to have four thirds squared all over two. Then minus, and then we're going to plug in our one here. So plugging in one, we're going to have one squared over two, which would just end up being one half. You can check this on a calculator, and all of this should come out to 28.

So 28 foot-pounds is the solution to the problem, and that is the amount of work that we have done when stretching the spring from 12 inches to 16 inches. So this is how you can solve work problems when it comes to springs. And pay very close attention to the units because typically, you want to have pounds and feet that you're dealing with or you want to have newtons and meters that you're dealing with. So if you ever have different units, just make sure to run the proper conversions to make sure that you get those right. Because recall that if you have inches instead, there are 12 inches in one foot. Now recall that if you have something like centimeters instead, there are 100 centimeters in one meter.

Typically, you want feet to be with pounds, and you want meters to be with newtons. This is going to be your force, and this right here would be the distance. So that's just something you want to keep in mind when you're solving these types of problems. So hope you found this video helpful. This was our solution for work down here, and let me know if you have any questions.

Let's move on.

concept

Lifting Problems

Video duration:

Video transcript

In recent videos, we've been spending a lot of time talking about work. And recall that when we want to find the work for the force acting on an object, we can simply integrate the force function from some distance a to b. And we've seen this before when it comes to these more basic types of work problems, but it turns out this equation and concept works every single time that you are dealing with a work problem, including more complicated scenarios. And in this video, we're going to be taking a look at these lifting problems, which is a more complicated example of dealing with work. Now I'm going to show you the steps for solving an example like this to hopefully make this difficult concept as straightforward as possible.

So let's jump right in. Now let's say we have an example like this. We're told a firefighter holds a 20 meter long rope with a density of three kilograms per meter from the top of the 20 meter tall building. What we're asked to do is find the work required to lift the rope a height of 10 meters at a constant speed. So this is a lifting problem, since we can see that we're asked to lift a rope.

Now, when it comes to these types of problems, there's something that you need to know. If you're trying to find the work to lift an object, the force in these examples is equivalent to the weight of the object that you're lifting. Now you want to pay close attention, because it's very easy to make a mistake if you're looking at the problem in general. So if you have a lifting problem like this, what some students might do is say, well, if we could find the weight of this rope, we could simply multiply that by the distances lifted, and that would give us our work. Now, this technically would be true if we were dealing with a constant force.

However, in this problem, the force is not constant. The reason why is because this rope here is being pulled up by the firefighter, but after a while, that rope is going to be pulled up a certain distance. Notice that the weight of the remaining rope is not going to be the same, because there's less and less rope remaining as the rope continues to be pulled. So we can't say that there's a constant weight or a constant force that we're applying here, the force is changing. So because the force is changing, we need to find some function for our force that we can integrate.

Thankfully, there is a way to do that, and that's using the equation that we see right here. So if you can find the density of the rope, or the weight per length, you can multiply that by the length function, and integrating this would be equivalent to integrating the force function. So doing this would be the same idea as we've seen for previous examples when dealing with work. So let's see if we can solve this example now. Now a good first step to take would be sketching a diagram or picture of the problem.

Now as we can see, there's a nice diagram given to us already that includes all the values we see in this example. So we can say this first step here is checked off. Now, the second step would be finding the weight per length of the rope, also the rope density. Now, keep in mind, there are a couple of things that you should be aware of when it comes to finding the weight density of the rope. You may be given your units in pounds per feet.

You also may be given your units in kilograms per meter, and the units are actually quite important. If you're given pounds per feet, this is a weight per length. That's a density right there for weight. So what you can do is take whatever that value is and plug it directly into the integral that we have. Now, alternatively, you might be given kilograms per meter, this is a mass per length, and because we have mass per length, we want to have a weight per length.

So what you need to do is take your mass in kilograms here and multiply it by 9.8, which is Earth's gravitational constant, its constant for acceleration. So you take whatever your value is, multiply it by 9.8, and that would give you your weight density. Now I can clearly see in the example we have here that we are given a mass density since it's in kilograms per meter. So what I need to do is take that three that I see and multiply it by 9.8 ms-2. So that's going to be the units for Earth's gravitational acceleration.

And then multiplying three times 9.8, well that's going to give me 29.4. So this right here would be the weight per length of our rope, and that would be step two for solving this example. So now that we've found this weight density right here, we now need to find the length. And the length is going to be a function of y, but how exactly do we find this function? Well, think about the situation we have over here.

Notice that we're lifting a rope. And as this firefighter continues to pull up the rope here, eventually it's going to be a certain length, say, right about there. Once we get to there, notice that the remaining piece of the rope here that we have is going to be this total length of the rope, minus however much we've lifted it, giving you this remaining length. So we can say that a function for our length function here is always just going to be the length of the rope minus y, whatever arbitrary height we're lifting the rope. So the function is going to be l minus y, where we can see l is 20, so specifically it's going to be 20 minus y as the function we're looking for.

So now we've found our length function. So we solve step three here, and our next piece is going to be determining the bounds on our integral. Now the bounds are going to go from zero all the way up to h. So the lower bound will always be zero for this, so that's going to be the lower bound, and h is going to be the height that the rope is lifted. Now in the example, we were actually given how far the rope is lifted.

We were told that the rope is going to be lifted 10 meters. So we're going to go from zero all the way up to 10. So we have our bounds, zero to 10, and now we can plug in all the information that we need to into this integral. So it's going to be the rope weight per length, the weight density, which is 29.4. And then it's going to be multiplied by our length function, which is 20 minus y.

From here, you just want to integrate this whole thing with respect to y, and that's going to give you the solution to this problem. So let's see how we can do this. Now what I'm going to do is take this 29.4, which is a constant, and bring it outside the integral. So we're going to have 29.4 multiplied by the integral from zero to 10 of 20 minus y. Now at this point, I know the rules for integration.

I can use the fundamental theorem of calculus to solve this. So what I can do is find the antiderivative of everything that I see in these parentheses. So that's all integrated with respect to y. So we're going to have 29.4. That's going to be multiplied by, and then the antiderivative of 20, that's going to be 20 y."

The antiderivative of y using the power rule is y22, and then this whole thing is going to be bounded from zero to 10. Now, at this point, you'll notice that zero is just going to set everything to zero inside of these brackets. So I really only need to worry about the 10 right here. So you can put 10 in for y everywhere that you see it, and from there, you can just do all the math and crunch these numbers. You can check them on a calculator, and you should get an answer of 4,410 joules."

So this is going to be the work done to pull up this rope, a height of 10 meters, and that is the solution to the problem. So, as you can see, you just need to go through these steps in this general process to find the pieces of this integral that we need. So we need to find the weight density and then the function length, and then we can plug that all in and integrate to get our answer. Now it's also possible you'll be asked some additional questions about the lifting scenario. And something like that may look like this example down here.

So notice that we found this work, but let's say that we were told that there's now an eight kilogram bucket attached to the end of the rope above. What is the total work required to lift both the rope and the bucket this same height? So in this example, we're now assuming we have the same situation we had up here, except now there is a bucket attached to the end of this rope. Now we're told this bucket is eight kilograms, and there's something that we know. The work to lift this rope, we already calculated to be 4,410 joules.

Now I want to specify. This right here is the work that we do for the rope, because we know that this is how we lifted the rope up to find this work. And we can also calculate the work to lift the bucket. Now this is not going to take as much effort on our part, because notice the bucket here actually would be a constant force this time. The weight of the bucket is going to be the same here as it is up here.

That's not going to change. So we can actually just use the equation for a constant force, which is going to be force times distance. Now the force that is caused by the bucket, well that's going to be the weight of the bucket, and we can calculate weight by taking whatever our mass is if it's in kilograms and multiplying it by 9.8. If it was in pounds, we could directly plug it in. But since it's in kilograms, we need to take our eight kilogram mass, multiply it by 9.8, Earth's gravitational acceleration, and then multiply that by the distance we're lifting it, which is 10 meters.

So we have eight times 9.8 times 10. And multiplying these numbers should give you seven eighty-four, and that's going to be in joules of work. So we now have the work required to lift the rope, which we figured out before, and the work required to lift the bucket. And if we want to find the work to lift both, which is what we're looking for in this problem, simply add the numbers together. So the total work that we're doing is going to be the work for the rope, 4,410 joules, plus the work for the bucket, 784 joules.

And adding these numbers will give you 5,194 joules, which is the solution to this problem. So this is how you can find the work required to lift the rope and the bucket. So, as you can see, what we're really doing here is just figuring out the total work of the situation that we have. If we were dealing with just the rope, since more and more rope is being pulled up, we have a variable force that we need to integrate. But since we had a bucket here where the weight is going to stay the same at every point that we lift the rope, then we need to just use the constant force, and then you want to add the two together if you want to find the total work required to lift both.

So this is the process for dealing with work. Hope you found this video helpful, and let's move on and get some more practice.

Problem

A $120 m$ chain hangs freely from the side of a building. The chain weighs $15 kg$ / $m$ . How much work is done to pull $80 m$ of the chain to the top of the building?

$940800 J$

$9600 J$

$4800 J$

$470400 J$

Problem

Find the work done by fully winding up a cable of length $30 ft$ and weight-density $2 lb$ / $ft$ .

$9000 J$

$0 J$

$840 J$

$900 J$

Problem

A $40 m$ rope hangs freely over a ledge. The density of the rope is $12 kg$ / $m$ . If a $5 kg$ bucket is attached to the end of the rope, how much work is done to pull the rope and the bucket to the ledge?

$94080 J$

$1960 J$

$96040 J$

$92120 J$

Problem

A $60 ft$ cable is attached to a cylinder that is attached to a winch. If the cable weighs 300 lbs, how much work is needed to wind $20 ft$ of the cable onto the cylinder using the winch? Hint: Divide cable weight by cable length to get density.

$3000 J$

$17900 J$

$5000 J$

$1700 J$

concept

Pumping Liquids

Video duration:

Video transcript

In recent videos, we've been spending a lot of time talking about how to solve work problems. Now, the general strategy for solving any kind of work problem is to take your force function and integrate it over a certain distance from a to b. We've seen this before, and it turns out some of these work problems can get quite complicated. In this video, we're going to take a look at one of the most complicated types of work problems, which, in my opinion, is dealing with pumping liquids. I will warn you, these problems do look quite challenging and can seem a bit abstract at first.

But pay close attention, because I'm going to walk you through the steps to solving these problems, as well as show you the general pattern to hopefully make these problems as intuitive as possible when you come across them. So let's get into things. Let's say we have this example down here, which tells us that a cylindrical tank of radius two feet and a height of 12 feet is eight feet full of oil that weighs 57 pounds per cubic foot. Since we're given this measurement in pounds per cubic foot, that means we're dealing with a weight density. We're asked to set up the integral to find the work required to pump the oil to the top of the tank.

This is clearly a pumping liquids problem, since I can see we're trying to pump oil to the top of the tank. A good first step when it comes to solving any type of pumping liquids problem would be to sketch a picture of the tank and the liquid. I can see in this problem, we're actually given a nice illustration of our tank right here. Notice we have this tank with the same measurements mentioned in the example. And since we can see we have the cylindrical tank, and it's filled with oil all the way up to eight feet, that is going to be the first step of this problem.

The question now becomes, what is the general strategy to finding the work required to pump this liquid to the top of the tank? Recall that when it comes to lifting problems, we really focus on two main things: the weight of the object and the distance that the object is lifted. We can take a similar approach when it comes to pumping liquids. Notice something, that's kind of what we're doing in this problem too. We're in essence just doing a more complicated lifting problem. We're trying to lift this liquid to the top of the tank, so we're forcing it that certain distance. However, it's not going to be as simple as just plugging in a force and a distance here because notice that each portion of liquid is traveling a different distance to get to the top of the tank.

We're also not given a direct weight for this liquid either. But there is a way that we can figure this out. You can take the density of your liquid, which we are often given, multiply it by the area of one cross section of that liquid, which we can figure out, and then multiply that by the distance. Since the distance is going to change depending on which portion of liquid we're looking at, we're going to need to find a function for our distance. Integrating all of this with respect to y would in essence be just integrating our force function over a certain distance, and that's going to allow us to find the work required to pump this oil to the top of the tank.

To find this density, there are a couple of things that you should know. If your density is given in pounds per cubic foot, that's a weight density that you can just directly plug into your integral. Now if your density is given in kilograms per cubic meter, that's a mass density. You need to multiply that by 9.8, which is Earth's gravitational acceleration, 9.8 meters per second squared, to convert this to a weight density.

But since I can see our problem clearly gives us our density in pounds per cubic foot, I can see that 57 is just going to be the density for this problem. The cross section of liquid here is going to be the same no matter where we look in the tank. It's the same circular cross section each time, so we can just find the area of this chunk right here. Since it's a circle with radius two, the area of a circle is pi r squared. So finding the area right here is going to be pi times two squared, which is the same thing as four pi. So that's going to be our third step.

Now we need to find our distance. Notice, if we're looking at this first portion of liquid right here, this top portion, we need to lift it this distance, which is going to be this top of the tank here minus that liquid level, which would be a distance of four feet. But notice if we were looking at this bottom portion right here, that would need to travel the entire distance to the top of the tank. So it'd be the top of the tank minus this value right here, which would just be 12 feet, the height that this needs to travel.

If we want to find a function for our distance, it's going to be the length of the tank, the height, minus y, some variable that changes depending on the height of the liquid. You can see here that the top of the tank is at 12 feet. So we're going to have our length of 12 minus y, and that's going to be our distance. We have our area, our distance, and now we have everything we need to put into this integral. The lower bound is usually going to be zero because this is with reference to the bottom of the tank. The upper bound is going to be the top layer of liquid. I can see the liquid level goes up to eight, so that means eight feet is going to be the top bound. We don't go all the way to the top of the tank because this liquid only covers a certain portion, and that's what we're integrating.

Now we have the density, which is 57, multiplied by the area, which is four pi, multiplied by this distance right here, which is 12 minus y. So this is everything that we need to integrate. Once you integrate with respect to y, the solution you should get is about 45,842 joules of work. So this right here should be the approximate solution to the problem, and that's how you can find the work required to pump a liquid to the top of a tank.

Keep in mind when you're dealing with these types of problems, the cross-sectional area depended on the area of a circle, which we figured out right here. But sometimes, you might have a different shape like a square or a triangle depending on the shape of the tank.

Keep very close attention to the details of the example and what the shape and overall structure of the tank is going to look like. Your area might be a function that you get rather than just this constant circle depending on whether or not that area changes. We'll look at some more examples to see these types of varying scenarios, so definitely try the examples and practice after this video. I hope you found this video helpful to really introduce you to the idea of finding the work required to pump liquids to the top of the tank.

See you in the next one.

example

Pumping Liquids Example 2

Video duration:

11m

Video transcript

So here we're told a tank in the shape of an inverted cone is 10 meters tall and has a radius of four meters. If the tank is full of gasoline, how much work is required to empty the tank? So, we need to take all this gasoline in the tank and pump it out of the top of the tank. How can we go about doing that? So this is a pumping liquids problem, and I am going to warn you, this one's going to be pretty tedious.

There's going to be a lot of steps that we need to take, but pay close attention, because I'm going to walk you through how we can solve this problem, and hopefully, you're going to see how we can solve many problems like this in the future, no matter how complicated or scary they look. So let's get into things. Now, the main thing you need to do with these types of problems is to apply this equation right here: The work is going to be equal to this integral of our density times area times distance. Now we'll go ahead and start with our first step here, which is going to be to draw a nice sketch or picture of the situation.

Now I can see in this problem we do have a nice sketch given to us, although keep something in mind here. This tank is all the way full of gasoline. So that means that this liquid is going to be filling this entire tank here. So we have a tank that's completely full of gasoline, and we need to pump all of it out. So we can go ahead and check off this first step here to see we have this all sketched out, and notice we also have this point right here, which is given to us, and we also have this radius, as well as just the general dimensions and shape of the tank.

Now our next step is going to be finding the density of the liquid. So here's the portion where we're actually starting to fill out this equation right here. Once we have our diagram, what I can do is try to find the density. And the way that we can do this is by seeing if it's given to us in the problem. Now keep in mind that if you're given the density in pounds per cubic feet, you want to plug in your density directly.

But if you're given kilograms per cubic meter, you want to multiply your result by 9.8 first. Now here's what's interesting. We actually are given the density of the problem, but notice it's in 6,670 Newtons per cubic meter. Is this something where we need to multiply it by 9.8, or can we plug it in directly? Well, it's not in pounds per cubic feet, but it's also not in kilograms per cubic meter.

This is in newtons per cubic meter. This is a force per volume. And because of that, well, we can say that this is a weight density. It's a weight force per volume. So this is a weight density.

Oftentimes, you're going to be given kilograms per cubic meters, which is a mass per volume, meaning you would need to multiply by 9.8 to get it to newtons. But since the newtons are already given to us here, we can see that the density is clearly shown in the problem. So the density, which I'm going to label as this Greek symbol rho here, this is oftentimes something you see for density. You may see this in your textbooks, depending on, depending on what your textbook uses. But in this case, I'm just going to say our density here is 6,670 newtons per cubic meter.

So we have our weight density, and that's our step two. Now step three is going to be finding the area of a cross-section. Now this will oftentimes be a function with respect to y. So what is the cross-section? Well, we can find the cross-section by looking at one slice of this fluid, one slice of the gas.

And if I'm looking at this, we look at one slice right here, I can see that we're going to get a circle that forms. But this is where things get a little interesting. Because even though we do have a circle that forms right here, this would be our radius. Now we know that when it comes to dealing with these types of circles, that the area is equal to pi r squared. But, in this case, there's not a direct radius I can plug in.

We could plug in the four here, but notice that's just the radius for the top of the tank right here. So it's not going to be the radius at every single point the fluid travels. It's not like in previous problems we've seen where we're dealing with a cylinder, which has the same circular radius throughout the shape. This is constantly changing. So what I need to do is take this radius and make it a variable, let's just say x.

And I'm making this x because notice it's changing along the x-axis right here that we've defined. So that's going to be x, and if our radius is pi r squared, that means that our area that we're going to deal with in this equation is going to be pi times this radius x squared. So we now have the area of this inverted cone, and we're getting closer to filling out this equation that we're trying to solve for. Now next, what I need to do is find the distance, which is going to be the length of the tank minus y. Now I can see here that the length of the tank, the tank top, shows up at 10.

So if I want to find our distance equation, well, the distance, which I'm going to go ahead and label as d here, that's going to be the length to the top right here, which is 10, minus y, because y is just this depth of the liquid. And as we continue to go further and further into this gas, notice that this number is going to get bigger, meaning it's going to take away more from the top of the tank. Eventually, when we get all the way to the bottom of the tank, there's going to be no more fluid left. So this is a function that's going to work for us. So now we have everything we need to plug into this equation.

So, from here what I need to do is determine the bounds, that's going to be my next step. Is to determine the bounds I'm actually going to write out this integral, so that way we can go ahead and have this. So using the integral up here, we're going to have that our work is equal to the integral, and I'll deal with the bounds in a moment here, but let's plug in everything we have. We have the density of 6,670 newtons per cubic meter. We have the area of pi times x squared, and then we have the distance, 10 minus y, and all of this is being integrated with respect to y.

Now how do I determine my bounds? Well, my bounds are going to go from the lowest part of the tank, which is usually going to be at zero, and we can see here it is zero, and then we're going to go all the way up to the top portion of the liquid. Now this tank is completely filled with gas. It's filled all the way to the top. This is just one small cross-section we're looking at for reference, but the problem clearly states that the tank is full of this gas.

So because of that, we're going to go all the way to the top of the tank, which is 10, so we're integrating what is the entire shape of this tank, the entire height. So that's going to be our setup for our integral. Now what I'm going to do is I'm going to try to shift some things around here to make this integral look a little more nice. So what I'm going to do is take this 6,670 and this pi, I'm going to pull them outside the integral since I know that those are constants. So we'll have that, and then the integral from zero to 10, and then we're going to have x squared times 10 minus y, all integrated with respect to y.

Now, at this point, what we need to do is integrate with respect to y. Now This might seem like a pretty straightforward step since we've learned a lot about integrals, but there's a bit of a problem you might notice here. We have an x, and we're integrating with respect to y. How do we deal with something like this? Well, what we need to do is somehow force there to be y, right here.

We need to get everything in terms of one variable. And there actually is a way that we can do that. Notice this portion of the tank is a line that we're looking at. And notice this x is going to change and vary as we go down or up, depending on this line that we're with reference to. So what I can do is find the equation for this line, which I know the equation for any line is y equals mx plus b.

This is the slope intercept form. Now what I can see here, first is my slope, and my slope is going to be rise over run. I can see that we rise 10 on this line, and then we run over four, since that's the radius at the top of the tank. So our slope is going to be 10 over four, which is the same thing as five over two. So we're going to have that y is equal to five over two times x plus b, which is the the y intercept.

Now looking at the y axis right here, I can see our intercept happens right here at y equals zero. So you can just say that b is equal to zero, so we don't even need to worry about writing it. So y equals five halves x, that's going to be what we can use for this equation. Now what I'm trying to do is get this so we have this x here, and that is isolated on one side, so everything is in terms of y. So to do that, what I'm going to do is multiply both sides of this equation by two fifths.

So two fifths multiplied on the right side and the left side, and that will allow me to cancel the two fifths and the five halves right there, leaving me with just x. And then we're going to have x is equal to two fifths y. And notice we now have an equation for x in terms of this y right here. So what that means is I can go down to my equation here and replace that x with two fifths y. So writing this all out again, we're going to have 6,670 pi multiplied by the integral from zero to 10, and then this is going to be times two fifths y squared times 10 minus y, all integrated with respect to y.

Now at this point, we can keep on going. So we have the 6,670 pi, and then two fifths squared, well, we can do that because two squared is four, five squared is 25, and we have the y squared right here. Then we're going to have 10 minus y all integrated with respect to y. Now at this point, what I can do is take this four twenty-fifths and multiply it by 6,670. Multiplying those numbers should give you 1,067.2, and we're going to multiply that by the integral from zero to 10 of y squared times 10 minus y, all integrated with respect to y.

Now we're going to keep on going here because we have this, and also there's going to be this pi right here upfront too. We can't forget about that. And so now continuing on this integral, I can distribute the y squared into these parentheses. So 1,067.2 times pi times the integral from zero to 10, and that's going to be 10 y squared minus y squared times y, which is y cubed, all integrated with respect to y. I'm going to go ahead and continue things up here since I'm running low on space.

So we're going to have that work is equal to, and now we can actually go ahead and apply the antiderivative. So we have 1,067.2 pi, and then what I can do is apply the antiderivative to this portion. So the antiderivative of 10 y squared using the power rule is going to be 10 y cubed over three. The antiderivative of y cubed is going to be y to the fourth power over four, and this is all going to be bounded from zero to 10. Now notice if I took zero and plugged it in for y, that would just set everything equal to zero.

So we only really need to worry about our 10 being plugged in here. So what the work is going to end up being is 1,067.2 times pi, and pi is about 3.14, and that's going to be multiplied by 10 times 10 cubed over three minus 10 to the fourth power over four, then all of that and that's everything plugged in here. So you just need to evaluate all these numbers on a calculator as you see them. And plugging in these numbers and multiplying, you should get your work approximately equal to 2,793,923 joules, and that is the work and the solution to this problem. So this is how you can solve these problems where you need to find the work for pumping liquids out of a tank.

Hope you found this video helpful, and let's move on and get some more practice.

example

Pumping Liquids Example 3

Video duration:

Video transcript

A spherical tank with an inner radius of nine feet is two-thirds full of oil that weighs 50 pounds per cubic foot. Find the work required to pump the oil through a hole in the top of the tank. Okay. So we have another pumping liquids problem, and as you know, these problems can get quite tricky. So there's going to be a lot of steps to solving this, but pay close attention, because I'm going to walk you through it to show you how you can deal with these situations when you come across them in your course.

So let's get right into things. Now, a good first step with any pumping liquids problem is to sketch a picture of the tank or the liquid. It really should be a combination of the two to really get the best diagram that you can. And I can see that we have a nice picture sketched for us right here. Notice we have a spherical tank.

We have our radius drawn here. We have the top layer of the oil, which is at 12. We have the top of the tank, and then we have the bottom of the tank right here. Our top layer of oil is at 12 because we're told that this is two-thirds full of oil. Two-thirds of 18 is 12, so that's how we got that number.

As you can see, we have sketched this picture of the tank, meaning this first step is checked off. Now, our next step would be finding the density of the liquid. Now there's a reason we want the density, and that's because this is the equation that's going to allow us to solve the problem. This is what you want to know if it comes to solving these types of pumping liquids problems, so we're just going to see if we can find our density, our area, and our distance, so we can plug it all in here and then integrate this function. Now we're starting with the density, and the density of the liquid may be given in pounds per cubic foot or kilograms per cubic meter, or it's also possible you'll see some other units, but it's always going to be something over a volume.

The question is whether it's a weight or mass. If it's in pounds or newtons, it's a weight. If it's in kilograms, it's a mass, which means you need to multiply by 9.8 first to get it into a weight because we want a weight density. Now I can see in our problem, we're given pounds per cubic foot. That is a weight density, so we can plug in this value directly.

So, the density, which I'm going to label as this Greek symbol rho here, this is how I like to label density, that's going to be 50. And that was just in pounds per cubic foot, so we have our density given to us. Now, next, what we need to do is find the area of a cross section. That's going to be our third step. Now, how do we go about doing that?

What I need to do is go somewhere in my tank and find a cross-sectional area. If I do this, notice that it's a circle that forms within the tank. Now, let's think about how the circle is going to change as we go up the tank or down the tank. We'll notice that our circles are either going to be larger or smaller depending on where we draw them in the tank. So that means this radius here is constantly changing.

Because of that, I'm going to say that this radius is just the variable x. I know the area of any circle is pi times the radius squared. So the area of this cross-section is going to be pi times this changing radius, x squared. And that radius changes depending on where you're whether you're close to the bottom, closer to the top, or somewhere in the middle. It's going to be largest at the very middle of this tank where we have the radius of nine.

So that's going to be our area, and that's step three. Now, step four is going to be finding the distance function, which is going to be the length minus y. Now this one's pretty straightforward. So if we're looking for our distance function, well, I can see the top of the tank is at 18. Then we can subtract off y, which would be this general distance for however far we are to get to the top of the tank.

So whatever that is, we'll call that y, and then this is going to be our distance. So that's step four. So now we've solved steps one through four. And then our fifth step will be determining the bounds. Now to go ahead and implement this fifth step, I'm going to take this equation and set it up.

So we're starting to get to a point where we can actually plug things into this equation. So we'll have the work is equal to the integral. I'll deal with the bounds in a moment here, of our density, which we figured out was 50, then times our area, which we figured out was pi x squared. That's going to be multiplied by our distance, which I can see is 18 minus y, and all of that is integrated with respect to y. Now as for the bounds, which is what we're trying to figure out, well, that can be figured out by taking a look at the tank and first starting with the bottom of the tank, which is usually going to be at zero.

I can see that is the case here, so zero will be my lower bound, and then my upper bound is going to be the top layer of the liquid, which we figured out was this top layer of 12. So we're going to go from zero to 12, and that's going to be our setup for the integral. So our last step would be to go ahead and just integrate this with respect to y, so we're in the home stretch for this problem, so let's see how we can do this. So to first start, I'm going to take this 50 pi here when starting to integrate, and pull it outside of the integral since I know that's a constant. Our bounds will stay from zero to 12, and then we're going to have x squared times 18 minus y, all integrated with respect to y.

Now this is where the problem gets especially tricky because notice we have an x right here, and we can't integrate x with respect to y. So we need to get x somehow in terms of y. But how do we go about doing that? What we need to do is look at our scenario right here, and think about how we could get x in terms of y. This is kind of sneaky looking at this at first, but there is a way to do things.

What we can do is notice that this is a circle right here, and this cross-section is going to follow the curve of this circle on both sides. Now the equation for a circle is going to be x squared plus y squared is equal to r squared. And I actually know what this radius is for the whole circle on the outer shape here, it's a radius of nine. So I can say x squared plus y squared equals nine squared. Now I know nine squared is 81, so I have x squared plus y squared equals 81.

But what am I trying to solve for here? Well, I'm ultimately trying to solve for x because that's going to get me in terms of this y right here. So to solve for x, I can just subtract this y squared on both sides of the equation. That'll get the y squareds to cancel right about there, giving me that x squared is equal to 81 minus y squared. And I can just take the square root on both sides of this equation, giving me that x is equal to the square root of 81 minus y squared.

So now we have x here in terms of our y variable. So what that means is if this is our x, going back to our area equation up here, our area would have actually been looking at this x here, it would have been pi times the square root of 81 minus y squared squared, which would just be 81 minus y squared. So how would this change in our equation down here? Well, that means I just need to take this x squared and replace it with this 81 minus y squared here, because that would be our radius that is being squared of our changing circle because its size is changing as we go up and down this tank. So what that means is we're going to get 50 pi times the integral from zero to 12 of and then we're going to have 81 minus y squared in for this x squared, because that's what we just figured out up here.

That's what the 81 minus y squared is. Then we're going to have multiplied by this portion, 18 minus y, all integrated with respect to y. Now we're in a much better position because notice everything here is in terms of y, and we're integrating with respect to y. Now we still have a little bit of a frustrating situation to deal with because we have to actually combine these into one full polynomial because trying to just do this as these two separate portions being multiplied is going to be a bit tricky.

So let's go ahead and do that. So we're going to have 50 pi times the integral from zero to 12, and then I'll multiply 81 by 18. Now 81 times 18, that's going to be 1,458. Now I need to multiply 81 by negative y to give me negative 81 y. I'm going to take negative y squared times 18 to give me minus 18 y squared.

We're just putting together this polynomial here. Then we're going to have plus, we'll have this negative y squared.replaceAll('д', '') times negative y, which is going to give me positive y cubed. And this is what we're integrating with respect to y. So now we have everything in terms of y, and we have this expanded out polynomial, so let's see how we can go about solving this. So we're going to have 50 pi, and then I need to take the antiderivative of all of these terms one at a time.

The antiderivative of this will be 1,458 y, then the antiderivative here, 81 y squared over two using that power rule. Then we'll take the antiderivative here, which you'll be 18 y cubed over three, plus and then we'll take this integral or antiderivative here, which would be y to the fourth over four, and then all of that is going to be bounded from zero all the way up to 12. Okay. Now if I go ahead and take my zero and I plug it in, notice that every place that we see y is multiplied by every term here. So that's just going to set everything to zero.

So I only need to worry about my top bound, which is 12. So doing this, we're going to have 50 pi is going to be that, and then we're going to have times and then we're going to have 1,458 times 12 minus 81 times 12 squared over two divided by three plus then we're going to have 12 to the fourth power over four. So those are going to be all the numbers that we need to crunch. And keep in mind that pi is about 3.14, so that's going to be your pi there. So you can multiply everything across the board here, and you should get that your work comes out to 1,017,876 joules, and that would be the solution to this problem.

So that is how you can find the work required to pump this liquid out of the top of this spherical tank. So that's the solution to the problem. That's how we can find the work required to pump this oil out. Hope you found this video helpful, and let's move on and get some more practice.

Problem

A water trough for horses has a triangular cross section with a height of $3 m$ and horizontal side lengths of $2 m$ . The length of the trough is $12 m$ . How much work is required to pump the water to the top of the trough when it is half full.

$176 comma 400 J$

$1 comma 764 comma 000 J$

$264 comma 600 J$

$88 comma 200 J$

Problem

A swimming pool has the shape of a rectangular prism with abase that measures 30 $ft$ by 20 $ft$ and is 5 $ft$ deep. The top of the pool is 1 $ft$ above the surface of the water. How much work is required to pump all the water out? Assume the density of water is 62.4 $lb$ / ${ft}^{3}$ .

449280 $ft ∙ lb s$

42120 $\operatorname{ft}\bullet\operatorname{lb}s$

79560 $\operatorname{ft}\bullet\operatorname{lb}s$

280800 $\operatorname{ft}\bullet\operatorname{lb}s$

example

Pumping Liquids Example 5

Video duration:

11m

Video transcript

A cylindrical tank with a length of 12 feet and a radius of three feet lays on its side and is full of molasses. We are asked to set up the integral to find the work required to empty the tank through an eight-foot-tall pipe at the top. So, there's a bit going on in this problem, and I want you to notice something.

We are not given a graph of the situation, nor steps or any recall equations. We just have to do this right from scratch. A good first step when dealing with any pumping liquids problem is to start by drawing a diagram, and that's what we're going to do. We're told it's a cylindrical tank, so it's going to look something like this: Here we have this tank that is on its side. We are told the radius is three feet, which means this distance right here is three feet, and it's 12 feet long. So, we have the length and radius of the tank, and we are told that we are pumping liquids through an eight-foot-tall pipe at the top of the tank.

Typically in such problems, you might see us draw these on a graph. I am going to draw this one on a graph as well, but this is where this problem actually becomes interesting. Typically, how we've done this in the past is by taking our shape and drawing it so that the bottom of the tank corresponds to resting on the x-axis. But this is one of the exceptions. This is a type of problem where it's actually going to be better if we draw it so that the origin of the graph corresponds with the center of the tank.

Let’s look at a front view cross-section here: It's just going to look like a circle, and rather than drawing it so the bottom rests on the x-axis, I’m going to draw it so it's now centered on the graph. This means the center point is right here, and since our radius is three feet, we need to go down three units and up three units from the center. This is what our graph's going to look like.

The tank is filled with molasses, which has this density; the entire tank is full, and we want to pump all this liquid out through the eight-foot-tall pipe on the top of the tank. To calculate the work, we need to determine everything that goes inside the integral for work. Recall there's an equation when dealing with pumping liquids: it's going to be the weight density of the fluid multiplied by some cross sectional area times the distance, and then that's all integrated with respect to y. Let’s start by looking for the density which we're given is a hundred pounds per cubic foot. Since we have pounds per cubic foot, we can just keep it as a hundred.

Next, we need to find a cross sectional area, which can be tricky because you want it to be a function of y. For this, we take a slice out of our cylinder, and this rectangle forms. It depends on the length of the tank, and its width varies. We recognize that the width where our slice goes through would be twice the x distance from the origin. This width right here is equivalent to two x, so the area would be 12⁢2⁢x.

To find what this x is, as a function of y, we apply the Pythagorean theorem since our distance from the center to the edge of the tank is the constant radius of three feet. So, <mrow> <msup><mi>x</mi><mn>2</mn></msup> <mo>+</mo> <msup><mi>y</mi><mn>2</mn></msup> <mo>=</mo> <msup><mi>3</mi><mn>2</mn></msup> </mrow> simplifies to <msqrt> <msup><mi>3</mi><mn>2</mn></msup> <mo>-</mo> <msup><mi>y</mi><mn>2</mn></msup> </msqrt>. Substituting this into the area expression, we get that the area is 24⁢.

Finally, the distance each slice needs to travel is 11⁢-⁢y, because of the eight-foot pipe plus three feet above the x-axis to the top of the tank, minus the variable y. Integrating, <mrow> <mn>100</mn> <mo>⁢⁢⁢, with bounds from -3 to 3.

This setup of the integral is what we’re asked to find. If you were to evaluate this, the result would be approximately 373,221 foot-pounds. But, following the problem's instructions, setting up the integral is all that was needed. This example shows why orienting the tank based on the graph's center can be crucial for simplifying calculations using basic geometric relations like the Pythagorean theorem. Changing orientations might make the problem much easier to solve mathematically. Feel free to ask any other questions or move on in the tutorial as needed.

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Here’s what students ask on this topic:

How do you calculate work done by a constant force in calculus?

To calculate work done by a constant force in calculus, you use the formula: $W = F d$ , where $F$ is the constant force applied, and $d$ is the distance over which the force is applied. The work is the product of the force and the distance, and it is measured in joules if the force is in newtons and the distance is in meters. This formula is applicable when the force is constant and acts in the direction of the displacement. If the force is not constant, you would need to use integration to find the work done.

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What is Hooke's Law and how is it used in work problems involving springs?

Hooke's Law states that the force required to stretch or compress a spring is proportional to the distance it is stretched or compressed. Mathematically, it is expressed as $F = k x$ , where $k$ is the spring constant, and $x$ is the displacement from the equilibrium position. In work problems, Hooke's Law is used to determine the force function, which is then integrated over the distance the spring is stretched or compressed to find the work done. The work done on a spring is given by the integral $\int_{0} k^{x} dx$ , which results in $\frac{1}{2} k x^{2}$ .

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How do you solve work problems involving lifting objects with variable force?

To solve work problems involving lifting objects with variable force, you need to account for the changing force as the object is lifted. The force is equivalent to the weight of the object, which may vary if the object is not uniform or if parts of it are lifted separately. You set up an integral of the force function over the distance the object is lifted. The integral is expressed as $\int_{a} b^{F} dy$ , where $F$ is the force function and $y$ is the variable representing the height. This approach allows you to calculate the total work done by integrating the variable force over the specified distance.

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What is the process for solving work problems involving pumping liquids?

Solving work problems involving pumping liquids requires calculating the work needed to move the liquid to a certain height. First, determine the weight density of the liquid, which is often given in pounds per cubic foot or kilograms per cubic meter. Then, find the cross-sectional area of the liquid being pumped. The work done is calculated by integrating the product of the weight density, cross-sectional area, and the distance each portion of liquid travels. The integral is set up as $\int_{0} h^{ρ} A (h - y) dy$ , where $ρ$ is the density, $A$ is the area, and $h$ is the height. This method accounts for the varying distances different portions of the liquid must travel.

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How do you determine the work done in lifting a rope with varying weight?

To determine the work done in lifting a rope with varying weight, you need to consider the changing weight as the rope is lifted. The force is not constant because the length of the rope being lifted decreases as more rope is pulled up. First, find the weight density of the rope, which is the weight per unit length. Then, set up an integral of the weight density multiplied by the length of the rope remaining to be lifted. The integral is expressed as $\int_{0} h^{ρ} (L - y) dy$ , where $ρ$ is the weight density, $L$ is the total length of the rope, and $y$ is the height. This approach allows you to calculate the total work done by integrating the variable force over the specified distance.

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Work: Videos & Practice Problems

Introduction To Work

Video transcript

How much work is required to push a chair across the floor with a force of F=8lb⁡F=8\operatorname{lb} from x=6ft⁡x=6\operatorname{ft} to x=9ft⁡x=9\operatorname{ft} along the xx-axis?

How much work is done by a person lifting a 10lb⁡10\operatorname{lb} bucket 4ft⁡4\operatorname{ft} off the ground?

How much work is required to move an object with a force of F(x)=4x3NF\left(x\right)=4x^3N acting along the xx-axis from x=0mx=0m to x=2mx=2m?

Compute the work done by a force of F=3x2NF=\frac{3}{x^2}N from x=2mx=2m to x=6mx=6m.

Work Done On A Spring (Hooke's Law)

Video transcript

A spring requires 12J12J of work to stretch the spring from 1.1m1.1m to 1.4m1.4m past its equilibrium. What is the spring constant?

A spring requires a force of 8N8N to stretch the spring to 5cm⁡5\operatorname{cm} past its equilibrium point. How much work could it take to stretch the spring from 2cm⁡2\operatorname{cm} to 9cm⁡9\operatorname{cm} past equilibrium?

Suppose a force of 10N10N is required to stretch a spring 0.5m0.5m from its equilibrium position. How much work is required to compress the spring 0.2m0.2m from its equilibrium position?

Work Done On A Spring (Hooke's Law) Example 1

Video transcript

Lifting Problems

Video transcript

A 120m⁡120\operatorname{m} chain hangs freely from the side of a building. The chain weighs 15kg⁡15\operatorname{kg}/mm. How much work is done to pull 80m⁡80\operatorname{m} of the chain to the top of the building?

Find the work done by fully winding up a cable of length 30ft⁡30\operatorname{ft} and weight-density 2lb⁡2\operatorname{lb}/ft⁡\operatorname{ft}.

A 40m⁡40\operatorname{m} rope hangs freely over a ledge. The density of the rope is 12kg⁡12\operatorname{kg}/mm. If a 5kg⁡5\operatorname{kg} bucket is attached to the end of the rope, how much work is done to pull the rope and the bucket to the ledge?

A 60ft⁡60\operatorname{ft} cable is attached to a cylinder that is attached to a winch. If the cable weighs 300 lbs, how much work is needed to wind 20ft⁡20\operatorname{ft} of the cable onto the cylinder using the winch? Hint: Divide cable weight by cable length to get density.

Pumping Liquids

Video transcript

Pumping Liquids Example 2

Video transcript

Pumping Liquids Example 3

Video transcript

A water trough for horses has a triangular cross section with a height of 3m3m and horizontal side lengths of 2m2m. The length of the trough is 12m12m. How much work is required to pump the water to the top of the trough when it is half full.

Pumping Liquids Example 5

Video transcript

Do you want more practice?

Here’s what students ask on this topic:

How do you calculate work done by a constant force in calculus?

What is Hooke's Law and how is it used in work problems involving springs?

How do you solve work problems involving lifting objects with variable force?

What is the process for solving work problems involving pumping liquids?

How do you determine the work done in lifting a rope with varying weight?

Your Calculus tutors

How much work is required to push a chair across the floor with a force of $F equals 8 lb$ from $x equals 6 ft$ to $x equals 9 ft$ along the $x$ -axis?

How much work is done by a person lifting a $10 lb$ bucket $4 ft$ off the ground?

How much work is required to move an object with a force of $F of x equals 4 x cubed N$ acting along the $x$ -axis from $x equals 0 m$ to $x equals 2 m$ ?

Compute the work done by a force of $F = \frac{3}{x^{2}} N$ from $x equals 2 m$ to $x equals 6 m$ .

A spring requires $12 J$ of work to stretch the spring from $1.1 m$ to $1.4 m$ past its equilibrium. What is the spring constant?

A spring requires a force of $8 N$ to stretch the spring to $5 cm$ past its equilibrium point. How much work could it take to stretch the spring from $2 cm$ to $9 cm$ past equilibrium?

Suppose a force of $10 N$ is required to stretch a spring $0.5 m$ from its equilibrium position. How much work is required to compress the spring $0.2 m$ from its equilibrium position?

A $120 m$ chain hangs freely from the side of a building. The chain weighs $15 kg$ / $m$ . How much work is done to pull $80 m$ of the chain to the top of the building?

Find the work done by fully winding up a cable of length $30 ft$ and weight-density $2 lb$ / $ft$ .

A $40 m$ rope hangs freely over a ledge. The density of the rope is $12 kg$ / $m$ . If a $5 kg$ bucket is attached to the end of the rope, how much work is done to pull the rope and the bucket to the ledge?

A $60 ft$ cable is attached to a cylinder that is attached to a winch. If the cable weighs 300 lbs, how much work is needed to wind $20 ft$ of the cable onto the cylinder using the winch? Hint: Divide cable weight by cable length to get density.

A water trough for horses has a triangular cross section with a height of $3 m$ and horizontal side lengths of $2 m$ . The length of the trough is $12 m$ . How much work is required to pump the water to the top of the trough when it is half full.