Work Done By Friction on a Circular Ramp

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SUMMARY

The work done by friction on a block moving down a rough quarter-circle ramp is calculated using the equation W = -(1/2)mv^2 + mgR, where m is the mass of the block, g is the acceleration due to gravity, V is the speed at the bottom, and R is the radius of the ramp. The gravitational potential energy at the top, mgR, converts to kinetic energy at the bottom, resulting in the derived equation for work done by friction. The discussion confirms that the calculations and reasoning presented are correct.

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  • Knowledge of friction and its effects on motion
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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) = [tex]\Delta[/tex]K + [tex]\Delta[/tex]U

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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