Calculating Work done by Friction

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Homework Help Overview

The discussion revolves around calculating the work done by friction on a block sliding down an inclined ramp. The problem involves a block of mass 18.0 kg on a 6.0-meter ramp inclined at 56.0 degrees, with a coefficient of kinetic friction of 0.48.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the work done by gravity and friction but is unsure about the correctness of their approach. Some participants question the use of cosine in the calculations and discuss the definition of work done by friction. Others suggest checking the work-energy theorem to verify the results.

Discussion Status

Participants are exploring different interpretations of the problem and discussing the calculations involved. There is no explicit consensus on the correct approach, but some guidance has been offered regarding the definitions and relationships between forces and work.

Contextual Notes

The original poster mentions constraints related to the number of attempts allowed for the question, indicating a pressure to arrive at a correct answer within limited tries.

reaperkid
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Homework Statement



Alrighty, I have most of this set but I'm getting the wrong answer and I'm not sure why.

A block of mass 18.0 kg is sliding down an 6.0 metre long ramp inclined at 56.0 deg. to the horizontal. If the coefficient of kinetic friction between the ramp and the block is 0.48, how much work is done by friction as the block moves from the top to the bottom of the ramp ?

m = 18kg
d = 6 m
Theta = 56 degrees
mu = .48

Homework Equations



W = Fd
F = ma
mu = fk / fn

The Attempt at a Solution



First I calculated work done by gravity. . .
W(g) = Fd = (18)(9.8)(6)(cos 56) = 591.85 J

Then I attempted to calculate the work done by friction. . .
W(f) = (.48*18*9.8*6*sin56) = 421.178 J

I tried that as a negative number as well, since the distance is negative.

So, clearly I'm missing something.

Any help would be greatly appreciated. Thanks!
 
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fk = mu*fn
And fn = mg*cos56
 
rl.bhat said:
fk = mu*fn
And fn = mg*cos56

Oh, should I be using cosine then??
 
I think your solution seems right. By definition the work done by friction is W = f (dot) d and f = muN = mu mgcos(theta). So W = fdcos(180deg). Its 180 degrees because f and d are in opposite directions. Then like you have it W = - fd = mu d mgcos(theta). It should be negative I think.

The only thing left to check that i can think of is if the work energy theorem is satisfied you can check that Wnet = the change in kinetic energy. I think you can do that by finding the final velocity of the object through kinematics. Wnet = Wg + Wf. If that checks then you can make sure that your answer is correct.

I hope this helps a little or makes it more clear...
 
Last edited:
Hells_Kitchen said:
I think your solution seems right. By definition the work done by friction is W = f (dot) d and f = muN = mu mgsin(theta). So W = fdcos(180deg). Its 180 degrees because f and d are in opposite directions. Then like you have it W = - fd = mu d mgsin(theta). It should be negative I think.

The only thing left to check that i can think of is if the work energy theorem is satisfied you can check that Wnet = the change in kinetic energy. I think you can do that by finding the final velocity of the object through kinematics. Wnet = Wg + Wf. If that checks then you can make sure that your answer is correct.

I hope this helps a little or makes it more clear...

Hmm, I don't get it.

Oh, well I just used my last guess. (It's online, we get 5 tries per question to get credit)

Thanks for the responses though!
 

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