Explanation Given: The mass of the bullet, m b = 20 g . The velocity of bullet, v b = 400 m/s The mass of a stationary block, m B = 2 kg The coefficient of kinetic friction between the block and the plane, μ k = 0.2 The acceleration due to gravity, g = 9.81 m/s 2 Show the free body diagram of the block as in Figure (1). Apply the conservation of linear momentum. m b v b + m B v B = ( m b + m B ) v (I) Here, the final velocity of a stationary block is v , initial velocity of a stationary block is v B . Apply the principle of impulse and momentum equation for y component. m ( v 1 ) y + ∑ ∫ t 1 t 2 F y d t = m ( v 2 ) y m ( v 1 ) y + N t − ( m b + m B ) g t = m ( v 2 ) y (II) Here, the initial and final velocity of the stationary block in y component is ( v 1 ) y and ( v 2 ) y respectively, initial and final linear momentum of the stationary block in y component is m ( v 1 ) y and m ( v 2 ) y , impulses acting on the stationary block from time limits t 1 to t 2 is ∑ ∫ t 1 t 2 F d t , and variable in time is dt . Apply the principle of impulse and momentum equation for x component. m ( v 1 ) x + ∑ ∫ t 1 t 2 F x d t = m ( v 2 ) x ( m b + m B ) ( v 1 ) x + [ − μ k N t ] = ( m b + m B ) v (III) Here, the coefficient of kinetic friction between the block and the surface is μ k , the initial and final velocity of the wooden block in x component is ( v 1 ) x and ( v 2 ) x respectively, initial and final linear momentum of the wooden block in x component is m ( v 1 ) x and m ( v 2 ) x , and impulses acting on the wooden block from time limits t 1 to t 2 is ∑ ∫ t 1 t 2 F d t . Determine the displacement of the block by integrating d s = v d t with the initial conditions s = 0 at t = 0. ∫ 0 s d s = ∫ 0 t v d t (IV) Conclusion: Substitute 20 g for m b , 0 for v B , 2 kg for m B , and 400 m/s for v b in Equation (I). ( 20 g ) ( 400 m/s ) + ( 2 kg ) ( 0 ) = ( 20 g + 2 kg ) v ( 20 g × 10 − 3 kg g ) ( 400 m/s ) = ( 20 g × 10 − 3 kg g + 2 kg ) v 0

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Author: HIBBELER

Publisher: PEARSON

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Chapter 15, Problem 43P

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The sliding distance of the stationary block before the bullet stops.

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Chapter 15, Problem 42P

Chapter 15, Problem 44P

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The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic friction between the block and the plane is μk = 0.32. Determine the distance the block will slide before it stops. Express your answer to three significant figures and include the appropriate units.

The 56.28 kg crate is hoisted up the 0 = 27° incline by the pulley system and motor M. If the crate starts from rest and, by constant acceleration, attains a speed of 8.17 m/s after traveling 7.84 m along the plane, determine the supplied power to the motor if the crate has moved 8 m and the coefficient of kinetic friction between the plane and the crate is Hk = 0.3. Neglect friction along the plane. The motor has an efficiency of 0.691. M

The force F, acting in a constant direction on the 22-kg block, has a magnitude that varies with the position s of the block. When s = 0 the block is moving to the right at v = 6 m/s. The coefficient of kinetic friction between the block and surface is μk= 0.3. Determine how far the block must slide before its velocity becomes 15 m/sm/s. Express your answer to three significant figures and include the appropriate units. Hint 1for Part A. How to approach the problem Use the principle of work and energy for the particle, which states that the particle's initial kinetic energy plus the work done by all the forces acting on the particle as it moves from its initial to its final position is equal to the particle's final kinetic energy. Note that besides force FF shown in the figure, the friction force acts on the block during its motion. Hint 2for Part A. How to find the friction force The friction force is directly related to the normal reaction, which in turn equals…

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