A small block with mass 0.0475 kg slides in a vertical circle of radius 0.525 m
ID: 1446487 • Letter: A
Question
A small block with mass 0.0475 kg slides in a vertical circle of radius 0.525 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 0.690 N
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
here,
mass , m = 0.0475 kg
radius , r = 0.525 m
let the speed at a be vA
and b be vB
at point A
normal force = 4 N
m * vA^2/r + m*g = 4
0.0475 * ( vA^2 /0.525 + 9.8) = 4
vA = 6.25 m/s
at point B
normal force , N = 0.69 N
m*vB^2/r - m * g = 0.69
0.0475 * ( vB^2 /0.525 - 9.8) = 0.69
vB = 3.57 m/s
using work energy theorm
work done by friction , W = initial kinetic energy - final kinetic energy - potential energy gained
W = 0.5 * m * vA^2 - 0.5 * m * vB^2 - m * g * (2 * r)
W = 0.0475 * ( 0.5 * 6.25^2 - 0.5 * 3.57^2 - 9.8 * 2 * 0.525)
W = 0.14 J
the work done on the block by friction during the motion of the block from point A to point B is 0.14 J