Calculating Work done by Friction

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The discussion revolves around calculating the work done by friction on a block sliding down a ramp. The user has calculated the work done by gravity and friction but is unsure why their answer is incorrect. Key points include the correct use of the work formula, considering the angle of friction, and ensuring the work done by friction is negative due to opposing directions. The work-energy theorem is suggested as a method to verify the calculations by checking if the net work equals the change in kinetic energy. The user expresses frustration after exhausting their attempts to solve the problem.
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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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