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Work done by frictional forces

  • Thread starter Zebra91
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  • #1
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Homework Statement


A disc of mass m slides with zero initial velocity down an inclined plane set at an angle alpha to the horizontal; having traversed the distance l along the horizontal plane, the disc stops. Find the work performed by the friction forces over the whole distance, assuming the friction coefficient k for both inclined and horizontal planes.


Homework Equations





The Attempt at a Solution


I am confused because in the solution the don't count in the work of gravitational force. I thought that the summed work is equal to the change in energy, so from that I would derive that the frictional forces work is equal to the change in energy minus gravitational work
 

Answers and Replies

  • #2
vela
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The problem statement is a bit unclear to me. Is this what's happening? The disc starts at rest on an inclined plane, rolls to the bottom (without slipping?), and then moves a distance l on a horizontal surface. Or is the distance l the change in the disc's horizontal position from the beginning to where it comes to rest again? Are you given the initial height of the disc?

It would help if you explain what you did using equations. When you only use words, we can only assume you understand the basic concepts, like how to calculate work, what total mechanical energy is, etc., but quite often, the problem is that students have misconceptions about these basic concepts. When you write equations, it's much easier to spot where you're going astray.
 
  • #3
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you cannot apply energy conservation here because change in potential energy is equal negative of work done by internel conservative foces
friction is not conservtive
 
  • #4
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The distance traveled after coming down the slope is l. We don't know the height. All we know are m, v0=0, alpha, l and k - coefficient of friction.
If at the start point potential energy is 0, than Ep=-mgh at the end. But h is unknown. Since kinetic energy is 0 at both starting and end point, preservation of energy does not apply.
W=Wt+Wg
 
  • #5
vela
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OK, so you have the total energy E0 = K0+U0 at the top is 0 because both the kinetic energy K0 and the potential energy U0 are 0. At the end, you have the total energy E is equal to -mgh because the kinetic energy again is 0 and the potential energy is -mgh. So you have the change in energy ΔE is

ΔE = E - E0 = -mgh - 0 = -mgh.

Now what are you saying this should be equal to?
 

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