Explanation Given: The mass of the block A is m A = 15 kg . The initial velocity of the block A is ( v A ) 1 = 10 m/s . The mass of the block B is m B = 10 kg . The coefficient of the friction is μ k = 0.3 . The distance between the blocks A and B is s = 4 m . The stiffness of the spring is k = 1000 N/m . The coefficient of restitution is e = 0.6 . Draw the free body diagram of the block A as shown in Figure (1). Write the principle of work and energy. T 1 + ∑ U 1 − 2 = T 2 (I) Here, T 1 is the initial kinetic energy, ∑ U 1 − 2 is the summation of work done by all the forces and T 2 is the final kinetic energy. Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 (II) Here, m is the mass, v is the velocity and the suffixes 1 , 2 indicates the initial and final stages. Write the formula for coefficient of restitution ( e ) . e = ( v B ) 2 − ( v A ) 2 ( v A ) 1 − ( v B ) 1 (III) Here, v is the velocity, the suffixes A , B indicates the block A and the block B and the suffixes 1 , 2 indicates the initial and final stages. Write the principle of conservation energy. T 1 + V 1 = T 2 + V 2 (IV) Here, T 1 , T 2 is the initial and final kinetic energy and V 1 , V 2 is the initial and final potential energy. Write the formula for kinetic energy. T = 1 2 m v 2 (V) Here, the blocks A and B are at the datum only. Hence the gravitation potential energy V g = 0 and the total potential energy ( V ) is equal to the elastic potential energy ( V e ) . V = V e Write the formula for elastic potential energy. V = V e = 1 2 k s 2 (VI) Here, k is the stiffness of the spring and s is the change in length of the spring. Conclusion: According to principle of work and energy. Consider the block A only. At initial: Calculate the initial kinetic energy of the block A . Substitute 15 kg for m and 10 m/s for v in Equation (V). T 1 = 1 2 15 kg ( 10 m/s ) 2 = 750 J Calculate the total work done of the block A . Refer Figure (1). The weight of the block and the normal reaction the block have the same magnitude and acting opposite direction. Hence the work done these forces will becomes zero. Here, Friction is the only force doing work on the block. The total work done on the block A is ∑ U 1 − 2 = − F f s = − μ k N A s Substitute 0.3 for μ k , 15 ( 9.81 ) N for N A and 4 m for s . ∑ U 1 − 2 = − 0.3 × 15 ( 9.81 ) N × 4 m = − 176 .58 J Calculate the final kinetic energy of the block A . Substitute 15 kg for m and ( v A ) 1 for v in Equation (V). T 2 = 1 2 15 kg ( v A ) 1 2 = 7.5 ( v A ) 1 2 Calculate the velocity of the block A just before collision with the block B . Substitute 750 J for T 1 , − 176.58 J for ∑ U 1 − 2 and 7.5 ( v A ) 1 2 for T 2 in Equation (I). 750 J+ ( − 176.58 J ) = 7.5 ( v A ) 1 2 ( v A ) 1 2 = 573.42 7.5 ( v A ) 1 = 76.456 ( v A ) 1 = 8.7439 m/s According to law conservation of linear momentum: At Initial: (before collision) Calculate the initial momentum. The initial momentum is ∑ m v 1 = m A ( v A ) 1 + m B ( v B ) 1 When the block B is at rest, the velocity of the block is zero. Substitute 15 kg for m A , 10 kg for m B , 8.7439 m/s for ( v A ) 1 and 0 for ( v B ) 1 . ∑ m v 1 = ( 15 kg ) ( 8.7439 m/s ) + ( 2 kg ) ( 0 ) = 131

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ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Basic Terminology in Mechanics

The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its enviro…

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Chapter 15, Problem 61P

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The maximum compression of the spring due to the collision:

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Chapter 15, Problem 60P

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The maximum compression of the spring due to the collision:

Solution Summary: The author illustrates the principle of work and energy in the free body diagram of the block A.

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Chapter 15 Solutions