Explanation Given: The mass of the marbles is 5 g . Initially the marblesare at rest. The glass tube surface is smooth. Neglect the size of the can. Draw the given diagram and free body diagram of the ball as shown in Figure (1). From Figure (1). Write the formula for work done byweight of the marbles. U W = m g s (II) Here, m is the mass of the marble, g is the acceleration due to gravity and s is the displacement. Write the formula for kinetic energy ( T ) . T = 1 2 m v 2 (III) Here, m is the mass of the marbles and v is the velocity of the marbles. Write the formula for Principle of Work and Energy. T i n i t i a l + ∑ U = T f i n a l (IV) Here, T i n i t i a l is initial kinetic energy, T f i n a l is final kinetic energy and ∑ U 1 − 2 is total work done by all the forces on the marbles. Write the formula for time ( t ) of marbles when it reaches the can. s = s 0 + v 0 t + 1 2 g t 2 (V) Here, s is the vertical displacement from the point B to the top of the can, s 0 is the initial displacement, v 0 is initial velocity of the marble and g is the acceleration due to gravity. Write the formula for placement distance R of the can. R = s 0 + v 0 t (VI) Conclusion: Refer Figure (1). Calculate the work done byweight of the marbles. When the vertical displacement of the marble from the point A to point B is 3 m − 2 m = 1 m . Substitute 5 g for m , 9.81 m/s 2 for g and 1 m for s in Equation (II). U W ( A − B ) = 5 g × 1 kg 1000 g × 9.81 m/s 2 × 1 m = 0.04905 J Calculate the total work done ∑ U A − B . ∑ U A − B = U W ( A − B ) Substitute 0.04905 J for U W ( A − B ) . ∑ U A − B = 0.04905 J Calculate the initial and final kinetic energy of the marbles. At initial: The marbles is at rest. T A = 0 At final: Substitute T B for T f i n a l , 5 g for m and v B for v in Equation (III). T B = 1 2 × 5 g × 1 kg 1000 g × v B 2 = 0.0025 v B 2 Calculate the velocity of the marble at the point B . Substitute 0 for T i n i t i a l , 0.04905 J for ∑ U A − B and 0.0025 v B 2 for T f i n a l in Equation (IV). 0 + 0.04905 J = 0.0025 v B 2 0.0025 v B 2 = 0.04905 v B = 0.04905 0.0025 v B = 4

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.3, Problem 31P

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The placement distance R of the can and the speed of the marbles fall into the can.

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Chapter 14.3, Problem 30P

Chapter 14.3, Problem 32P

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Chapter 14 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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The placement distance R of the can and the speed of the marbles fall into the can.

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The placement distance R of the can and the speed of the marbles fall into the can.

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Chapter 14 Solutions