Explanation Given: The weight of the block A is 10 lb . The weight of the block B is 3 lb . Initial velocity of the block is ( v B ) 1 = 3 ft/s at t = 0 s . The coefficient of kinetic friction between the horizontal plane and block A is μ k = 0.15 . Draw the free body diagram of the system as shown in Figure (2). Write the principle of linear impulse and momentum. m v 1 + ∫ t 1 t 2 F d t = m v 2 (I) Here, m is the mass of the block, v is the velocity of the block, F is the force acting on the block, t is the time and the suffixes 1 , 2 indicates the initial and final stages. Conclusion: Refer Figure (1a). The total length of the cord is l = s A + 2 s B . The change length of the cord is 0 = δ s A + 2 δ s B δ s A = 2 δ s B Similarly the velocity of the blocks is v A = 2 v B At initial ( v B ) 1 = 3 ft/s . Substitute 3 ft/s for ( v B ) 1 . ( v A ) 1 = 2 ( 3 ft/s ) = 6 ft/s Consider block A only: Calculate the total impulse on the block A . Refer figure ( 1 b ) . Here, the weight of the block ( W ) is equal to the normal reaction ( N ) on the block. Hence the friction force acting on the block is F f = μ k N = 0.15 ( 10 lb ) = 1.5 lb . The net impulse on the block ( A ) due to the tension ( T ) and the friction F f is ∑ ∫ t 1 t 2 F d t = T ( t ) − F f ( t ) When t = 1 s , substitute 1 s for t and 1.5 lb for F f . ∑ ∫ t 1 t 2 F d t = T ( 1 s ) − 1.5 lb ( 1 s ) = T − 1.5 Calculate the final velocity of the block ( A ) . Substitute 10 lb for m , 6 ft/s for v 1 and T − 1.5 for ∑ ∫ t 1 t 2 F d t in Equation (I). 10 lb × 1 32.2 ft/s 2 ( 6 ft/s ) + T − 1.5 = 10 lb × 1 32

Engineering Mechanics: Dynamics (14th Edition)

14th Edition

ISBN: 9780133915389

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 15.2, Problem 32P

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The velocity of the block A at t=1 s .

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Chapter 15.2, Problem 31P

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The velocity of the block A at t = 1 s .

Solution Summary: The author illustrates the principle of linear impulse and momentum by drawing a free body diagram of the system.

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The velocity of the block A at t=1 s .

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