A small block with mass 0.0450kg slides in a vertical circle of radius 0.575m on
ID: 1264674 • Letter: A
Question
A small block with mass 0.0450kg 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 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 0.685N .
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.
Wfriction = ??Explanation / Answer
m = 0.0450 kg
r = 0.575 m
F1 = 3.85 N
F2 = 0.685 N
force acting on m at point A
F1 - mg = ma --------(1)
where g = 9.8 m/s^2 and a is acceleration
force acting on m at point B
mg - F2 = ma -------(2)
(1) + (2)
F1 - F2 = 2ma
or a = (F1 - F2/2m)
work done by friction
W = ma * r