A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m on
ID: 1967374 • Letter: A
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