A small block with mass 0.0425 kg slides in a vertical circle of radius 0.550 m
ID: 1344246 • Letter: A
Question
A small block with mass 0.0425 kg slides in a vertical circle of radius 0.550 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 3.95 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 0.665 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.Explanation / Answer
at point A
the total force is given by
Fc = m v2 / r + m g
3.95 = 0.0425v2 /0.550 + 0.0425 X 9.8
3.95 = 0.077v2 + 0.416
0.077v2 = 3.534
v2 = 3.534 / 0.077 = 45.89
v = 6.77 m/s
at Point B
Fc= m v2 / r - m g
0.665 = (0.0400 v2 / 0.550) - 0.0425 X 9.8
0.665 = 0.077 v2 - 0.4165
0.077 v2 = 1.0815
v2 = 1.0815 / 0.077
v2 = 14.045
v = 3.74 m/s
now the energy at point A
PE + KE = m g h + 1/2 m v2
= 0 +1/2 X 0.0425X 6.772
= 0.9739 J
now the energy at point B
PE + KE = m g h + 1/2 m v2
=0.0425 X 9.8 X 1.1 + 1/2 X 0.0425 X 3.742
= 0.7553 J
now the final energy is = 0.9739 - 0.7553
= 0.2186 J