Calculating Work done by Friction

  • Thread starter reaperkid
  • Start date
  • #1
14
0

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!
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
fk = mu*fn
And fn = mg*cos56
 
  • #3
14
0
fk = mu*fn
And fn = mg*cos56
Oh, should I be using cosine then??
 
  • #4
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:
  • #5
14
0
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!
 

Related Threads on Calculating Work done by Friction

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
11
Views
60K
Replies
16
Views
4K
Replies
2
Views
13K
Replies
8
Views
1K
Replies
2
Views
7K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
Top