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CP A small block with mass 0.0400 kg slides in a vertical
Chapter 9, Problem 78P(choose chapter or problem)
CP A small block with mass 0.0400 kg slides in a vertical circle of radius R = 0.500 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 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 normal force exerted on the block has magnitude 0.680 N. How much work is done on the block by friction during the motion of the block from point A to point B?
Questions & Answers
QUESTION:
CP A small block with mass 0.0400 kg slides in a vertical circle of radius R = 0.500 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 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 normal force exerted on the block has magnitude 0.680 N. How much work is done on the block by friction during the motion of the block from point A to point B?
ANSWER:Solution 78P Introduction We have to calculate total energy at the bottom and top. The difference in the total energy is the work done by friction. Step 1 The weight of the block is And the normal reaction at the bottom of the path is At the bottom the centrifugal force and the weight acts in the same direction. Hence the total force acting on the surface will be the sum of centrifugal force and the weight, and hence the normal reaction is the sum of the weight and centrifugal force. So the centrifugal force is So the kinetic energy of the particle at the bottom is