A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m on

Question

A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m 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 0.665N .
How much work was done on the block by friction during the motion of the block from pointA to point B?

Explanation / Answer

at bottom,
3.85 = 0.0350 x 9.8 + 0.0350 x v^2/ 0.525
Speed = v = 7.25 m/s

at top,
0.665 =0.0350 x u^2/0.525 - 0.0350 x 9.8
speed = u = 3.888 m/s
Use work energy theorem

From,

0.5mu^2 - 0.5mv^2 = -mg(2R) + Wf

Wf = -0.295 Joules

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