What is the total amount of work done by friction?

[OGBJ]

Homework Statement

Consider a rope of mass M and length L, resting on a horizontal table, as shown in the figure. The coefficient of kinetic friction between the table and the rope is μk. Let's consider the work that's done by friction as we slide the rope off the table.

a.) Consider a small section of rope. What is the mass dm of that small section. (You can use the following variables in your answer: M, L, dx)
ANSWER: dm = M(dx/L)

c.) What is the magnitude of the friction force acting on this small piece of rope? (You may use the following variables in your answer: M, L, dx, μk)
ANSWER: μkMg(dx/L)

d.) Calculate the small amount of work dW that the friction force does in sliding a small section of the rope dm initially located a distance x from the edge of the table off the table.
ANSWER: dW = -μkxMg(dx/L)e.) What is the total amount of work done by friction as the entire rope slides off the table?

Homework Equations

dW = (-μkxMg)(dx/L)

The Attempt at a Solution

I know I need to integrate both sides of the relevant equation to solve for W. I thought all the terms but x and dx could be treated as constants, so I took them out of the integral and was left with:
W = (-μkMg/L)∫xdx = (-μkx^2Mg)/(2L).
However, this is not the correct solution, which leads me to assume that I can't just take the other terms out of the equation like that. I'm not sure how to treat the problem otherwise, though. Any help would be appreciated. Thanks.

 

[haruspex]

= -μkx^2Mg/2L.

Is that exactly how you entered it in an online tool? Do you understand that what that says is "μ_kx²(Mg/2)L"?
Try using parentheses correctly.
Edit: and as berkeman noticed and I should have, x should not appear in the answer.

 

[berkeman]

Welcome to the PF.

ANSWER: dW = -μkxMg(dx/L)

Is this a correct answer, or just your answer so far? The amount of work to slide the mass dm off the table depends on the distance to the end of the table from that dm, not on dx...

EDIT -- Or maybe you just left a term off of this equation...?

 

[OGBJ]

Is that exactly how you entered it in an online tool? Do you understand that what that says is "μ_kx²(Mg/2)L"?
Try using parentheses correctly.

No. I believe I formatted it correctly in my attempted submission. Here's what it looks like in the tool:

(I think I failed to get the image to embed. Here's the a direct link to it if needed: https://imgur.com/a/ZE8iG)
Apologies for the incorrect formatting in the body of the post. I'll clean that up now.

 

[OGBJ]

Welcome to the PF.

Is this a correct answer, or just your answer so far? The amount of work to slide the mass dm off the table depends on the distance to the end of the table from that dm, not on dx...

EDIT -- Or maybe you just left a term off of this equation...?

Thanks. The answers to parts a)->d) have been entered into and accepted by WebAssign. Here's where I am so far; hopefully this clears things up: https://imgur.com/a/FFhZI

Edit: Why should x not appear in the answer? Shouldn't it be expected that work increases as distance does? Also, as per part a, the M(dx/L) portion of dw=mukxgM(dx/L) can be interpreted as dm, since that's what's solved for in part a. I'm not certain if that's relevant, but I thought I'd mention it, since I'm pretty confused about where I am in the problem right now.

 

[haruspex]

Thanks. The answers to parts a)->d) have been entered into and accepted by WebAssign. Here's where I am so far; hopefully this clears things up: https://imgur.com/a/FFhZI

Please see my edited post and reread berkeman's carefully. What is the range of the integral?

 

[OGBJ]

Please see my edited post and reread berkeman's carefully. What is the range of the integral?

I don't think I'm fully understanding why x shouldn't be in the answer. Here's my edit that I made just after you replied: Why should x not appear in the answer? Shouldn't it be expected that work increases as distance does? Also, as per part a, the M(dx/L) portion of dw=mukxgM(dx/L) can be interpreted as dm, since that's what's solved for in part a. I'm not certain if that's relevant, but I thought I'd mention it, since I'm pretty confused about where I am in the problem right now.

As for the range, I don't believe there should be one--should it not just be indefinite? If it was definite the only range I could see is 0->x.

Edit: If the range was 0->x nothing would change since when x = 0 the whole equation is equal to zero.

 

[berkeman]

I understand that the problem is trying to guide you to approach the problem in a particular way, but maybe solving it a different way will help you gain some intuition.

When I first saw the overall problem, I approached it this way... For each position of the rope as it slides off the table, you can calculate W(x) for a small displacement dx. In this case, x is the length of the rope still left on the table. So just write the equation for W(x) and integrate it from x=L to x=0 (to correspond to the rope being pulled off of the table). Maybe give that a try to see what the answer should look like.

 

[haruspex]

As for the range, I don't believe there should be one--should it not just be indefinite?

You are asked for the total work done as the rope of length L slides off, not some unknown length x of it.

 

[OGBJ]

I understand that the problem is trying to guide you to approach the problem in a particular way, but maybe solving it a different way will help you gain some intuition.

When I first saw the overall problem, I approached it this way... For each position of the rope as it slides off the table, you can calculate W(x) for a small displacement dx. In this case, x is the length of the rope still left on the table. So just write the equation for W(x) and integrate it from x=L to x=0 (to correspond to the rope being pulled off of the table). Maybe give that a try to see what the answer should look like.

Ah. Integrating using L as my limit rather than x solves my problem. -mukMgL/2 is my final answer! Thank you.