A small block with mass 0.0300 kg slides in a vertical circle of radius 0.450 m
ID: 1342517 • Letter: A
Question
A small block with mass 0.0300 kg slides in a vertical circle of radius 0.450 m 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 3.85 N . 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 0.690 N.
How much work was done on the block by friction during the motion of the block from point A to point B?
Explanation / Answer
Given,
mass 0.0300 kg
radius 0.450 m
track has magnitude 3.85 N
the block has magnitude 0.690 N
at bottom,
3.85 = 0.0300 x 9.8 + 0.0300 x v^2/ 0.450
Speed = v = 7.30 m/s
at top,
0.690 =0.0300 x u^2/0.450 - 0.0300 x 9.8
speed = u = 3.841 m/s
Use work energy theorem
From,
0.5mu^2 - 0.5mv^2 = -mg(2R) + Wf
Wf = -0.31355 Joules
work was done on the block by friction during the motion of the block from point A to point B is -0.31355 Joules
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