Explanation Given data: Two blocks are connected by a wire. Mass per unit length of the wire ( m L ) is 8.50 × 10 − 4 kg m . Consider the 42.0 kg block as block 1, and 19.0 kg block as block 2. Mass of block 1 m 1 is 42.0 kg. Mass of block 2 m 2 is 19.0 kg. The blocks are pulled across a horizontal frictionless floor by a horizontal force P → . The force is applied to less massive block (block 2). A transverse wave travels on the wire between the blocks. The speed of the wave ( v ) is 352 m s . The mass of the wire is negligible compared to the mass of the blocks. Formula used: Refer to the equation (16.2) in the textbook and write the expression for the speed of the wave on the wire between the blocks follows. v = F ( m L ) (1) Here, F is the tension in the wire between the block, ( m L ) is the mass per unit length of the wire between the blocks, which is given as 8.50 × 10 − 4 kg m , m is the mass of the wire between the block, and L is the length of the wire between the block. From the given data, draw the block diagram for the given description, and free-body diagram for each block as shown in Figure 1. In the Figure 1, F N1 is the normal force on the block 1, F N2 is the normal force on the block 2, P → is the pulling force to the blocks, which is to be determined, m 1 g is the weight of the block 1, and m 2 g is the weight of the block 2. From Figure 1, write the expression for the tension in the wire for block 1 as follows. F = m 1 a (2) Here, m 1 is the mass of block 1, a is the acceleration of the wave on the wire. From Figure 1, write the expression for the tension in the wire for block 2 as follows. P → − F = m 2 a (3) Here, m 2 is the mass of block 2. Rewrite the expression in equation (2) for the acceleration of the wave on the wire as follows. a = F m 1 Substitute ( F m 1 ) for a in equation (3), P → − F = m 2 ( F m 1 ) Rewrite the expression for the pulling force to the blocks as follows. P → = m 2 ( F m 1 ) + F P → = F ( m 2 m 1 + 1 ) (4) Take square on both side of the expression in equation (1)

10th Edition

ISBN: 9781118486894

Author: David Young, Shane Stadler

Publisher: WILEY

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Chapter 16, Problem 107AP

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Chapter 16, Problem 106AP

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