A small block with mass 0.0350 kg slides in a vertical circle of radius 0.400 m
ID: 1600507 • Letter: A
Question
A small block with mass 0.0350 kg slides in a vertical circle of radius 0.400 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 4.00 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 0.670 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
normal reaction at bottom
N = mg + mu^2/R
NR = mgR +mu^2
NR/2 = mgR/2 + initial kinetic energy
initial kinetic energy = 0.7314 J
normal reaction at bottom
N = mV^2/R + mg
NR/2 - mgR/2 = final kinetic energy
final kinetic energy = 0.0654 J
work done by friction = change in mechanical energy
Wf = change in kinetic energy + change in potential energy
Wf = -0.3916J
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.