A small block with mass 0.0375 slides in a vertical circle of radius 0.600 on th
ID: 2060770 • Letter: A
Question
A small block with mass 0.0375 slides in a vertical circle of radius 0.600 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 3.90 . 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 .
How much work was done on the block by friction during the motion of the block from point A to point B?
Explanation / Answer
At the bottom of the circle, the force exerted must equal the centripetal force plus the weight of the block,
So, 3.9 = mv2/r + mg
3.9 = (.0375)(v2)/.6 + (.0375)(9.8)
v = 7.52 m/s
Then we can figure out how much KE the block had at this point
KE = .5mv2 = (.5)(.0375)(7.52)2 = 1.06 J
At the top of the ramp, the force extered is the centripetal force minus the weight of the block,
So., .690 = (.0375)(v2)/.6 - (.0375)(9.8)
v = 4.11 m/s
The KE at the top is .5mv2 = (.5)(.0375)(4.11)2 = .317 J
The block also has PE at the top found by mgh
PE = (.0375)(9.8)(1.2) = .441 J
Total Energy at the top is .441 + .317 = .758 J
The amount of energy lost, is the work done by friction
W = (1.06 - .758) = .302 J