A small block with mass 0.0500kg slides in a vertical circle of radius 0.575m on

Question

A small block with mass 0.0500kg slides in a vertical circle of radius 0.575m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 4.00N . In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has

How much work was done on the block by friction during the motion of the block from point A to point B?

Express your answer with the appropriate units.

magnitude 0.680N .

Explanation / Answer

When the block is at the bottom ,

let the speed at the bottom is Vb

N = 4 N

4 = 0.05 * 9.8 + 0.05*vb^2/.575

vb = 6.35 m/s

Now , at the top , spped is vt

0.680 = 0.05 vt^2/.575 - 0.05 * 9.8

vt = 3.67 m/s

let the work done by friction is Wf

Now , using work energy theroum ,

- 0.5 * 0.05 * *(6.35^2 - 3.67^2) = -0.05 * 9.8 * 2 * .575 + wf

wf = - 0.108 J

work done by fricition is -0.108 J

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