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

this problem, we're asked to compute the work that's done given this force right here from a distance of x equals two to x equals six meters. So how do we go about solving this? Well, when solving these problems, we should first take a look at the force function. Notice we're given a function with x in it. That means that we're not dealing with a constant force in this problem. So whenever we have non constant forces that we're dealing with, what we need to do is take our force as a function of x here and integrate it to based on the distance. Now you can see the distance goes from two meters to six meters, so that means our bounds on the integral are gonna go from two all the way up to six. So what I can do is plug in my force function. So we're going to have two to six of our force function, which I can see right here is three over x squared, and we're integrating this with respect to x. Now from here, what I can do is take this constant three that I see and pull it outside of the integral. That's a rule that I should know about. So we have three, and we're integrating from two to six, of one over x squared with respect to x. Now how do I go about solving this integral right here? Well, to solve this, recall that one over x squared can actually be rewritten. We can say that this whole thing is the same thing as x to the negative two power. So we're gonna have the same integral, x to the negative two, that we integrate with respect to x. Now at this point, this is a power rule that we know how to solve. We can add one to the exponent and then divide by that new exponent. So if I do that, what I'm going to end up with is we'll have our three on the outside here, and it's going to be the antiderivative of this function, which comes out to x to the negative one over negative one. And then this whole thing is still bounded from two to six. Now at this point, I can simplify this a little bit more. So x to the negative one over negative one, that's the same thing as negative one over x. Then this whole thing is still bounded from two six. So now we've gotten to this point. Now I'm gonna go ahead and continue this over here. So we're trying to calculate work, and then what I can do is plug in my upper and lower bound. I'm gonna start with six. So doing this, we're gonna keep this three on the outside here, and then we're going to have six that we plug in for our function. So we're gonna get negative one over six. Then what I need to do is take this whole thing right here. I need to subtract off plugging in now two. So we're gonna put two into our function, which will give us and then I'll put this in brackets here. It's gonna give us negative one over two. So these are the results that we're subtracting. So we're gonna have these two results, and then this is going to be closed off right there. So what we need to do is combine negative one over six minus negative one over two, and minus a negative becomes plus a positive. So what we really need to do is combine this negative one over six plus one half. Now one half, that's gonna be the same thing as six or or three over six. So what I can do is remove this and say that this is the same thing as three over six. Now the reason I'm doing this is so we can get these common denominators right here. So combining these, we're gonna have this three on the outside here, and then we're going to have negative one sixth plus three sixth, which comes out to just two sixths. So we're going to have two sixths right there. Now two sixths is the same thing as one third, and then one third times three all just comes out to one. So our work is going to be one joule, and that is the solution to the problem. So this is how you can calculate work when you're given a force function that is not constant. So as you can see, we just integrated our function here. We applied our bounds, and that led us to our solution. Hope you found this video helpful, and let's move on.

10. Physics Applications of Integrals

Work

Calculus

10. Physics Applications of Integrals

Work

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Multiple Choice

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

$1 J$

$negative 0.12 J$

$0.36 J$

$0.67 J$

Verified step by step guidance

Step 1: Recall the formula for work done by a variable force: $ W = \int_{a}^{b} F(x) \, dx $, where $ F(x) $ is the force as a function of position $ x $, and $ a $ and $ b $ are the limits of integration.

Step 2: Substitute the given force $ F(x) = \frac{3}{x^2} $ and the limits $ x = 2 $ to $ x = 6 $ into the formula. This gives $ W = \int_{2}^{6} \frac{3}{x^2} \, dx $.

Step 3: Simplify the integrand $ \frac{3}{x^2} $ as $ 3x^{-2} $ to make it easier to integrate. The integral becomes $ W = \int_{2}^{6} 3x^{-2} \, dx $.

Step 4: Use the power rule for integration: $ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C $ (for $ n \neq -1 $). Applying this to $ 3x^{-2} $, the integral becomes $ W = 3 \left[ \frac{x^{-1}}{-1} \right]_{2}^{6} = -3 \left[ \frac{1}{x} \right]_{2}^{6} $.

Step 5: Evaluate the definite integral by substituting the limits $ x = 6 $ and $ x = 2 $. This gives $ W = -3 \left( \frac{1}{6} - \frac{1}{2} \right) $. Simplify the expression to find the work done.