(a) Explanation Given: The mass of the block is m 1 = 20 kg . The mass of the block is m 2 = 5.0 kg . Formula used: The Free body diagram for the blocks is shown in Figure 1 Figure 1 Consider the tension in the rope is T , acceleration of block 1 is a 1 and block 2 is a 2 . Let in some time, the block m 1 moves a certain distance and at the same time, the block m 2 descends double the distance. The relation between the acceleration of both the blocks is given by, a 2 = 2 a 1 The acceleration of block 1 by Newton’s law of motion is given by, m 1 a 1 = 2 T a 1 = 2 T m 1 The expression for the tension in the string is evaluated as m 2 g − T = m 2 a 2 m 2 a 2 = m 2 g − T m 2 ( 2 a 1 ) = m 2 g − T 2 m 2 a 1 = m 2 g − T Solve further as 2 m 2 ( 2 T m 1 ) = m 2 g − T m 2 g = 4 m 2 m 1 T + T T = m 2 g (b) To determine The tension in the string.

Physics for Scientists and Engineers, Vol. 1

6th Edition

ISBN: 9781429201322

Author: Paul A. Tipler, Gene Mosca

Publisher: Macmillan Higher Education

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Chapter 4, Problem 81P

(a)

To determine

The acceleration of each of the blocks.

(b)

To determine

The tension in the string.

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Chapter 4, Problem 80P

Chapter 4, Problem 82P

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The acceleration of each of the blocks.

Solution Summary: The free body diagram for the blocks is shown in Figure 1

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