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} / f t 3 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

Video Lessons 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

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Introduction To Work Video Summary

In the study of calculus, one important application is in solving work problems, which involve calculating the work done on an object when a force is applied over a distance. Work is defined as the product of the force applied to an object and the distance over which that force is applied. The unit of work is measured in joules when the force is in newtons and the distance is in meters. Alternatively, if the force is measured in pounds and the distance in feet, work is measured in foot-pounds.

To calculate work when a constant force is applied, the formula used is:

Work = Force × Distance

For example, if a constant force of 22 newtons is applied to move an object from x = 1 meter to x = 4 meters, the distance traveled is 3 meters. Thus, the work done can be calculated as:

Work = 22 N × 3 m = 66 joules

However, when dealing with a variable force, the approach changes. In this case, the work done is represented by the area under the force function curve. To find this area, we use integration. The work done can be expressed as:

Work = ∫_a^b F(x) dx

where F(x) is the force function, and a and b are the limits of integration representing the distance over which the force is applied.

For instance, if the force function is given by F(x) = x² and we want to find the work done from x = 1 to x = 4, we would set up the integral:

Work = ∫₁⁴ x² dx

To solve this, we first find the antiderivative of x², which is:

F(x) = (x³)/3

Next, we evaluate this from 1 to 4:

Work = [(4³)/3] - [(1³)/3] = (64/3) - (1/3) = 21 joules

In summary, the key takeaway is that the work done on an object is calculated as the area under the force curve. For constant forces, the simple multiplication of force and distance suffices, while variable forces require integration to determine the area under the curve. Understanding these concepts is essential for solving various real-world problems involving work in physics and engineering.

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)

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Work Done On A Spring (Hooke's Law) Video Summary

In calculus, the concept of work is defined as the integral of a force function over a distance. This means that to calculate work, we find the area under the curve of the force function using integration. A specific application of this concept is in problems involving springs, which are often analyzed using Hooke's Law. Hooke's Law states that the force exerted by a spring is proportional to its displacement from the equilibrium position, and is mathematically represented as:

$$ F(x) = k \cdot x $$

where $ F(x) $ is the force applied to the spring, $ k $ is the spring constant (measured in newtons per meter or pounds per foot), and $ x $ is the distance the spring is stretched or compressed from its equilibrium position.

The equilibrium point is crucial in these calculations, as it represents the position where the spring is neither compressed nor stretched. When a spring is stretched or compressed beyond this point, work is done on the spring, which can be calculated by integrating the force function over the distance moved. The work done on the spring when stretching it from point $ a $ to point $ b $ can be expressed as:

$$ W = \int_{a}^{b} F(x) \, dx $$

To illustrate this, consider a spring with a spring constant $ k = 3 \, \text{N/m} $. If we want to calculate the work done to stretch the spring from 4 meters to 6 meters, we first establish the force function:

$$ F(x) = 3x $$

Next, we set up the integral with the bounds from 4 to 6:

$$ W = \int_{4}^{6} 3x \, dx $$

Using the power rule for integration, we find:

$$ W = \left[ \frac{3x^2}{2} \right]_{4}^{6} $$

Calculating this gives:

$$ W = \frac{3(6^2)}{2} - \frac{3(4^2)}{2} = \frac{3(36)}{2} - \frac{3(16)}{2} = \frac{108}{2} - \frac{48}{2} = \frac{60}{2} = 30 \, \text{joules} $$

Thus, the work done to stretch the spring from 4 meters to 6 meters is 30 joules. This process highlights the importance of determining the force function and applying the work equation through integration to solve spring-related problems effectively.

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

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

In this example, we explore the application of Hooke's Law to determine the spring constant and the work done in stretching a spring. Hooke's Law states that the force exerted by a spring is proportional to the distance it is stretched from its natural length, expressed mathematically as:

F = k \cdot x

where F is the force applied, k is the spring constant, and x is the displacement from the equilibrium position.

In this scenario, the spring has a natural length of 8 inches, and it takes 24 pounds of force to stretch it to 12 inches. To find the spring constant k, we first determine the displacement:

x = 12 \text{ inches} - 8 \text{ inches} = 4 \text{ inches}

Next, we convert the displacement from inches to feet, knowing that 12 inches equals 1 foot:

x = \frac{4}{12} \text{ feet} = \frac{1}{3} \text{ feet}

Substituting the values into Hooke's Law gives us:

24 = k \cdot \frac{1}{3}

To solve for k, we multiply both sides by 3:

k = 24 \cdot 3 = 72 \text{ pounds per foot}

Now that we have the spring constant, we can calculate the work done in stretching the spring from 12 inches to 16 inches. The work done on a spring can be calculated using the integral of the force function:

W = \int_{a}^{b} F(x) \, dx

Here, the force function based on Hooke's Law is:

F(x) = k \cdot x = 72x

We need to determine the bounds for the integral. The lower bound corresponds to 12 inches (1 foot), and the upper bound corresponds to 16 inches, which we convert to feet:

16 \text{ inches} = \frac{16}{12} \text{ feet} = \frac{4}{3} \text{ feet}

Thus, the work done is:

W = \int_{1}^{\frac{4}{3}} 72x \, dx

Factoring out the constant, we have:

W = 72 \int_{1}^{\frac{4}{3}} x \, dx

The antiderivative of x is:

W = 72 \left[ \frac{x^2}{2} \right]_{1}^{\frac{4}{3}}

Evaluating this from 1 to 4/3 gives:

W = 72 \left( \frac{(\frac{4}{3})^2}{2} - \frac{1^2}{2} \right)

Calculating the squares:

W = 72 \left( \frac{\frac{16}{9}}{2} - \frac{1}{2} \right) = 72 \left( \frac{8}{9} - \frac{1}{2} \right)

Finding a common denominator (18) for the fractions gives:

W = 72 \left( \frac{16}{18} - \frac{9}{18} \right) = 72 \left( \frac{7}{18} \right)

Thus, the work done is:

W = 28 \text{ foot-pounds}

In summary, the spring constant is 72 pounds per foot, and the work required to stretch the spring from 12 inches to 16 inches is 28 foot-pounds. It is crucial to maintain consistent units throughout the calculations, typically using pounds and feet or newtons and meters, to ensure accurate results.

concept

Lifting Problems

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Lifting Problems Video Summary

Understanding work in physics involves integrating the force function over a distance. This principle applies universally, including in more complex scenarios like lifting problems. When lifting an object, the force corresponds to the weight of the object, but it’s crucial to recognize that this force can vary, especially when dealing with objects like ropes.

Consider a scenario where a firefighter lifts a 20-meter long rope with a density of 3 kilograms per meter from the top of a building. To find the work required to lift the rope a height of 10 meters at a constant speed, we first need to determine the weight density of the rope. Since the density is given in kilograms per meter, we convert this to weight per length by multiplying by the acceleration due to gravity, which is approximately 9.8 m/s². Thus, the weight density is calculated as:

Weight Density = Density × Gravitational Acceleration = 3 \, \text{kg/m} \times 9.8 \, \text{m/s}² = 29.4 \, \text{N/m}.

Next, we establish a function for the length of the rope being lifted. As the firefighter pulls the rope, the length of the remaining rope decreases. The length function can be expressed as:

Length Function: $ L(y) = L - y = 20 - y \, \text{meters}, $

where $ L $ is the total length of the rope and $ y $ is the height lifted.

To find the total work done, we set up an integral from 0 to the height lifted (10 meters) of the weight density multiplied by the length function:

Work = $ \int_0^{10} 29.4 \cdot (20 - y) \, dy. $

Factoring out the constant, we have:

Work = $ 29.4 \int_0^{10} (20 - y) \, dy. $

Calculating the integral, we find the antiderivative:

Antiderivative = $ 20y - \frac{y^2}{2} $ evaluated from 0 to 10.

Substituting the bounds, we compute:

Work = \( 29.4 \left[ (20 \cdot 10) - \frac{10^2}{2} \right] = 29.4 \left[ 200 - 50 \right] = 29.4 \cdot 150 = 4410 \, \text{Joules}. \

This result represents the work done to lift the rope. If an additional weight, such as an 8-kilogram bucket, is attached to the rope, we can calculate the work required to lift the bucket separately. The weight of the bucket is:

Weight = Mass × Gravitational Acceleration = 8 \, \text{kg} \times 9.8 \, \text{m/s}² = 78.4 \, \text{N}.

Since the bucket's weight remains constant during the lift, the work done to lift the bucket is:

Work = Force × Distance = 78.4 \, \text{N} \times 10 \, \text{m} = 784 \, \text{Joules}.

To find the total work required to lift both the rope and the bucket, we simply add the two amounts of work together:

Total Work = Work for Rope + Work for Bucket = 4410 \, \text{J} + 784 \, \text{J} = 5194 \, \text{Joules}.

This comprehensive approach illustrates how to tackle lifting problems involving variable forces through integration, while also addressing constant forces when applicable. Understanding these principles is essential for solving various work-related problems in physics.

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

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Pumping Liquids Video Summary

When solving work problems, particularly those involving pumping liquids, the general strategy involves integrating the force function over a specified distance. This approach can become complex, especially in scenarios like pumping oil from a tank. To illustrate this, consider a cylindrical tank with a radius of 2 feet and a height of 12 feet, filled with oil to a height of 8 feet. The oil has a weight density of 57 pounds per cubic foot, which is crucial for calculating the work required to pump it to the top of the tank.

The first step in tackling such problems is to visualize the scenario, often by sketching the tank and the liquid. This helps in understanding the dynamics of the situation. The work done in pumping the liquid can be thought of as a lifting problem, where the weight of the liquid and the distance it needs to be lifted are key factors. However, since different portions of the liquid will travel varying distances to reach the top, a more nuanced approach is necessary.

To find the work required, start by determining the density of the liquid, which in this case is already provided as 57 pounds per cubic foot. Next, calculate the cross-sectional area of the liquid. For a cylindrical tank, the area can be found using the formula for the area of a circle: $ A = \pi r^2 $. Here, with a radius of 2 feet, the area becomes $ A = 4\pi $ square feet.

Next, establish a function for the distance each portion of liquid must be lifted. The distance varies depending on the height of the liquid, represented by $ y $. The distance to lift the liquid to the top of the tank can be expressed as $ 12 - y $, where 12 feet is the total height of the tank.

With the density, area, and distance defined, the integral to calculate the work can be set up. The bounds of the integral will range from 0 to 8 feet, corresponding to the height of the liquid. The work integral can be expressed as:

\[ W = \int_{0}^{8} (57)(4\pi)(12 - y) \, dy \]

Upon evaluating this integral, the result will yield the total work required to pump the oil to the top of the tank, which is approximately 45,842 joules. It's important to note that the cross-sectional area may vary depending on the shape of the tank; thus, careful attention to the tank's geometry is essential when solving similar problems.

In summary, understanding the relationship between weight density, cross-sectional area, and distance is crucial in calculating the work needed to pump liquids. Practice with various tank shapes will enhance your problem-solving skills in this area.

example

Pumping Liquids Example 2

Video duration:

11m

Pumping Liquids Example 2 Video Summary

To determine the work required to empty a tank shaped like an inverted cone, we start by understanding the dimensions and properties of the tank. The tank is 10 meters tall with a radius of 4 meters and is filled with gasoline. The work done in pumping the gasoline out can be calculated using the formula for work:

$$ W = \int \rho \cdot A \cdot d $$

where $ W $ is the work, $ \rho $ is the density of the liquid, $ A $ is the area of the cross-section, and $ d $ is the distance the liquid must be pumped. In this case, the density of gasoline is given as 6,670 Newtons per cubic meter, which is a weight density.

Next, we need to find the area of a cross-section of the tank at a depth $ y $. The area $ A $ of a circle is given by:

$$ A = \pi r^2 $$

However, since the radius changes with depth in an inverted cone, we express the radius as a function of $ y $. The relationship between the radius $ r $ and the height $ y $ can be derived from the geometry of the cone. The slope of the line representing the radius as a function of height is determined by the dimensions of the cone:

$$ r = \frac{2}{5}y $$

Thus, the area becomes:

$$ A = \pi \left(\frac{2}{5}y\right)^2 = \frac{4\pi}{25}y^2 $$

The distance $ d $ that each slice of gasoline must be pumped to the top of the tank is given by:

$$ d = 10 - y $$

Now we can set up the integral for work:

$$ W = \int_0^{10} 6,670 \cdot \frac{4\pi}{25}y^2(10 - y) \, dy $$

Factoring out the constants gives:

$$ W = 6,670 \cdot \frac{4\pi}{25} \int_0^{10} y^2(10 - y) \, dy $$

Next, we expand the integrand:

$$ y^2(10 - y) = 10y^2 - y^3 $$

Now we can compute the integral:

$$ W = 6,670 \cdot \frac{4\pi}{25} \left( \int_0^{10} (10y^2 - y^3) \, dy \right) $$

Calculating the antiderivative:

$$ \int (10y^2 - y^3) \, dy = \frac{10}{3}y^3 - \frac{1}{4}y^4 $$

Evaluating from 0 to 10 gives:

$$ \left[ \frac{10}{3}(10)^3 - \frac{1}{4}(10)^4 \right] = \left[ \frac{10}{3}(1000) - \frac{1}{4}(10000) \right] = \left[ \frac{10000}{3} - 2500 \right] = \frac{10000 - 7500}{3} = \frac{2500}{3} $$

Substituting back into the work equation:

$$ W = 6,670 \cdot \frac{4\pi}{25} \cdot \frac{2500}{3} $$

Calculating this gives:

$$ W \approx 2,793,923 \text{ joules} $$

This result represents the total work required to pump all the gasoline out of the tank. Understanding the relationship between the geometry of the tank, the properties of the liquid, and the application of calculus is essential in solving such problems effectively.

example

Pumping Liquids Example 3

Video duration:

Pumping Liquids Example 3 Video Summary

To determine the work required to pump oil from a spherical tank, we start by understanding the tank's dimensions and the properties of the liquid. The tank has an inner radius of 9 feet and is filled with oil weighing 50 pounds per cubic foot, reaching a height of 12 feet, as it is two-thirds full.

First, we establish the density of the oil, denoted by the Greek letter ρ, which is given as 50 pounds per cubic foot. This value is crucial for calculating the weight of the oil being pumped.

Next, we need to find the cross-sectional area of the oil at a certain height y. The cross-section forms a circle, and its radius changes depending on the height. We denote this radius as x. The area A of the circle can be expressed as:

A = $\pi x^2$

To find x in terms of y, we use the equation of a circle:

$x^2 + y^2 = r^2$

Substituting the radius of the tank (9 feet), we have:

$x^2 + y^2 = 81$

Rearranging gives:

$x^2 = 81 - y^2$

Thus, the area becomes:

A = $\pi (81 - y^2)$

The distance the oil must be pumped to reach the top of the tank is given by:

$d = 18 - y$

Now, we can set up the integral to calculate the work W required to pump the oil from the tank:

$W = \int_0^{12} ρ \cdot A \cdot d \, dy$

Substituting the values we have:

$W = \int_0^{12} 50 \cdot \pi (81 - y^2)(18 - y) \, dy$

Next, we expand the integrand:

$W = 50\pi \int_0^{12} (1458 - 81y - 18y^2 + y^3) \, dy$

Calculating the antiderivative gives:

$W = 50\pi \left[ 1458y - \frac{81y^2}{2} - 6y^3 + \frac{y^4}{4} \right]_0^{12}$

Evaluating this from 0 to 12, we find:

$W = 50\pi \left[ 1458(12) - \frac{81(12^2)}{2} - 6(12^3) + \frac{12^4}{4} \right]$

After performing the calculations, we arrive at the total work required to pump the oil, which is approximately 1,017,876 joules. This comprehensive approach illustrates how to tackle problems involving the work done in pumping liquids from a tank.

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 $lb$ / $f t^{3}$ .

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

Pumping Liquids Example 5 Video Summary

To determine the work required to empty a cylindrical tank filled with molasses through an eight-foot tall pipe, we start by visualizing the problem. The tank has a length of 12 feet and a radius of 3 feet, positioned horizontally. The density of molasses is given as 100 pounds per cubic foot, which is a weight density.

First, we establish a coordinate system where the center of the tank is at the origin. This allows us to represent the tank as a circle in the xy-plane, with the equation:

$$x^2 + y^2 = 3^2$$

From this, we can derive the width of a slice of liquid at a height y, denoted as w. The width can be expressed as:

$$w = 2x = 2\sqrt{3^2 - y^2} = 2\sqrt{9 - y^2}$$

Thus, the cross-sectional area A of the slice is:

$$A(y) = 12 \cdot w = 12 \cdot 2\sqrt{9 - y^2} = 24\sqrt{9 - y^2}$$

Next, we calculate the distance each slice of liquid must be lifted to exit through the pipe. The total height to the top of the pipe is 11 feet (3 feet from the center of the tank to the top plus 8 feet of the pipe). Therefore, the distance d for each slice is:

$$d = 11 - y$$

Now, we can set up the integral for work W, which is the product of the weight density, the area, and the distance, integrated over the height of the liquid in the tank. The work done to lift each slice of liquid is given by:

$$W = \int_{-3}^{3} \text{(density)} \cdot A(y) \cdot d \, dy$$

Substituting the known values, we have:

$$W = \int_{-3}^{3} 100 \cdot 24\sqrt{9 - y^2} \cdot (11 - y) \, dy$$

To simplify, we can factor out the constants:

$$W = 2400 \int_{-3}^{3} \sqrt{9 - y^2} (11 - y) \, dy$$

This integral setup captures the work required to pump the molasses out of the tank through the pipe. The bounds of integration reflect the vertical extent of the liquid in the tank, from -3 feet to 3 feet, ensuring that we account for the entire volume of liquid being lifted.

Do you want more practice?

We have more practice problems on Work

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.

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 $\frac{1}{2} k x^{2}$ .

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.

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.

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

Introduction To Work Video Summary

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

How much work is done by a person lifting a 10lb10\operatorname{lb} bucket 4ft4\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)

Work Done On A Spring (Hooke's Law) Video Summary

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 5cm5\operatorname{cm} past its equilibrium point. How much work could it take to stretch the spring from 2cm2\operatorname{cm} to 9cm9\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

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

Lifting Problems

Lifting Problems Video Summary

A 120m120\operatorname{m} chain hangs freely from the side of a building. The chain weighs 15kg15\operatorname{kg}/mm. How much work is done to pull 80m80\operatorname{m} of the chain to the top of the building?

Find the work done by fully winding up a cable of length 30ft30\operatorname{ft} and weight-density 2lb2\operatorname{lb}/ft\operatorname{ft}.

A 40m40\operatorname{m} rope hangs freely over a ledge. The density of the rope is 12kg12\operatorname{kg}/mm. If a 5kg5\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 60ft60\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 20ft20\operatorname{ft} of the cable onto the cylinder using the winch? Hint: Divide cable weight by cable length to get density.

Pumping Liquids

Pumping Liquids Video Summary

Pumping Liquids Example 2

Pumping Liquids Example 2 Video Summary

Pumping Liquids Example 3

Pumping Liquids Example 3 Video Summary

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

Pumping Liquids Example 5 Video Summary

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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?

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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.