Work Done By Friction on a Circular Ramp

AI Thread Summary
The discussion revolves around calculating the work done by friction on a block sliding down a rough quarter-circle ramp. The work done by gravitational potential energy is expressed as mgR, which converts to kinetic energy at the bottom of the ramp. The derived equation for work done by friction is W = -(1/2)mv^2 + mgR. Participants confirm that this approach is correct, addressing initial doubts about the derivation process. The conversation highlights the importance of understanding energy transformations in physics problems.
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Homework Statement



A small block of mass m is held at the top of a rough quarter-circle ramp of radius R as shown in the figure, and then released from rest. When it reaches the bottom of the ramp, it is moving with speed V. How much work did friction do on the block from the top to the bottom? Express your answer in terms of m, g, V and R.

th_WorkDonebyFriction.jpg


Homework Equations



W (nc) = \DeltaK + \DeltaU

The Attempt at a Solution



Work done by gravitational potential energy is equal to mgR. It turns into kinetic energy at the bottom. Therefore,

W = ((1/2)mv^2 - 0) + (0 - mgR)
W = (1/2)mv^2 - mgR

And since this the work done by the object, the work done by friction would be the negative value. Thus,

W = -(1/2)mv^2 + mgR

Am I on the right track?
 
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Yes, that's actually the right answer. Why did you doubt that it was?
 
ideasrule said:
Yes, that's actually the right answer. Why did you doubt that it was?

Haha, I'm not sure. I was never really good at deriving equations.
 

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