Explanation Given information: Mass of block A is equal to 2 k g . Mass of ball B is 1 k g . Ball B is suspended from 1 m rope. Co-efficient of friction between block and surface is equal to 0.2. d = 1.5 m Co-efficient of restitution is equal to e = 0.8 N . α = 30 ° Tension in the rope is 20 N . The principle of conservation of energy is defined as: “When a particle moves under the action of conservation of forces, the sum of kinetic energy and potential energy of that particle remains constant” T 1 + V 1 = T 2 + V 2 Calculation: Apply Newton’s second law of motion: ↖ Σ F n = m B a n T − m B g c o s α = m B a n Rearrange, T = m B ( a n + g c o s α ) 20 N = ( 1.0 ) ( a n + 9.81 c o s 30 ° ) a n = 11.504 m / s 2 Now, find the speed of ball B: a n = v 2 2 ρ Substitute and solve, 11.504 m / s 2 = v 2 2 1 m v 2 = 3.392 m / s According to conservation of energy: T 1 = 1 2 m B ( v B 1 ) 2 V 1 = 0 T 2 = 1 2 m B v 2 2 V 2 = m B g h 2 = m B g l ( 1 − cos α ) Substitute, T 1 + V 1 = T 2 + V 2 Then, 1 2 m B ( v B 1 ) 2 + 0 = 1 2 m B ( 3.392 m / s ) 2 + m B ( 9.81 m / s 2 ) ( 1 m ) ( 1 − cos 30 ° ) Therefore, v B 1 = 3.759 m / s We know that, ( v B 1 ) n − ( v A 1 ) n = e [ ( v B ) n − ( v A ) n ] Therefore, v B 1 cos θ − v A 1 = e ( v A − 0 ) 3

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Conservation of Energy

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Chapter 13, Problem 13.200RP

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The initial compression x of the spring.

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Chapter 13, Problem 13.199RP

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Vector Mechanics for Engineers: Dynamics

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The initial compression x of the spring.

Solution Summary: The author explains Newton's second law of motion, which states that when a particle moves under the action of conservation of forces, the sum of kinetic energy and potential energy remains constant.

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The initial compression x of the spring.

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