Problem 7.78 A small block with mass 0.0500kg slides in a vertical circle of rad
ID: 1317901 • Letter: P
Question
Problem 7.78 A small block with mass 0.0500kg slides in a vertical circle of radius 0.500m 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 3.75N. 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 0.670N. Part A 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 bottom
Normal reaction = m*(v^2)/r + mg = 0.05*(v^2) / 0.5 + 0.05*9.8
0.05*(v^2) / 0.5 + 0.05*9.8 = 3.75
v (bottom) = 5.7096 m/s
at top
Normal reaction = m*(v^2)/r - mg = 0.05*(v^2) / 0.5 - 0.05*9.8
0.670 = 0.05*(v^2) / 0.5 - 0.05*9.8
v = 4.8166 m/s
now by energy conservation,
energy at bottom = energy at top
(1/2)*m*(v(bottom))^2 = (1/2)*m*(v(top))^2 + mg*(2*r) + work done by friction
work done by friction = 0.2610 Jules