A small block with mass 0.0450kg slides in a vertical circle of radius 0.600m on

Question

A small block with mass 0.0450kg slides in a vertical circle of radius 0.600m 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 3.85N. 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 magnitude 0.660N. How much work was done on the block by friction during the motion of the block from point A to point B?

Explanation / Answer

at bottom N-mg = mv2 /r

3.85-0.45 = (0.045/0.600 ) *V2

v = 6.73 m/sec

at top

N+mg = mv2 /r

0.660+0.45 = (0.045/0.600)*v2

v = 3.8 m/sec

now by work energy theorem

total work done = change in kinetic energy

W friction + Wgravity = KEbottom - KEtop

Wf - mg(2r) = 1/2 m *(3.8)2 - 1/2 m *(6.73)2

Wf =0.54 - 0.694 = -0.154 j

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