Explanation Given data: Mass of the child ( m ) is 20 kg. Length of the chain ( L ) is 3.0 m. Refer to the Figure P10.77 in the textbook for the swinging of a child. Angle made by the child before reversing the direction ( θ ) is 45 ° . From the figure, the final velocity (speed) of the child ( v f ) at the highest point of swing is 0 m s since the child loses complete velocity at the highest point and reverses her direction. Formula used: Write the expression for conservation of energy for the given conditions as follows. K i + ( U g ) i = K f + ( U g ) f (1) Here, K i is the initial kinetic energy (at the lowest point of the swing), ( U g ) i is the initial gravitational potential energy, K f is the final kinetic energy (at the highest point of the swing), and ( U g ) f is the final gravitational potential energy. Write the expression for kinetic energy. K = 1 2 m v 2 (2) Here, m is the mass of the object, and v is the velocity (speed) of the object. Write the expression for gravitational potential energy. U g = m g y (3) Here, g is the acceleration due to the gravity, which is 9.8 m s 2 , and y is the height. Explanation: Consider the air resistance, drag forces, and frictional affects in the system. Consider the chain as massless. Consider initial velocity of the child at the lowest point of the swing or bottom of the arc as v 0 , which is to be determined. Consider the final velocity of the child at the highest point of the swing as v 1 , which is 0 m s since the child loses the complete velocity at the highest point of her swing. Consider the initial height at the lowest point of the arc as y 0 , which is 0 m. Consider the final height of the child at the highest point of the swing as y 1 , which is calculated as follows. Refer the Figure P10.77 in the textbook and write the mathematical expression to find the final height as follows. y 1 = L − L cos θ = L ( 1 − cos θ ) Substitute 3.0 m for L , 45 ° for θ , y 1 = ( 3.0 m ) [ 1 − cos ( 45 ° ) ] = ( 3.0 m ) ( 1 − 0.7071 ) = 0.8787 m Calculation of initial kinetic energy: Modify the expression in equation (2) to find the initial kinetic energy of the child as follows. K i = 1 2 m ( v i ) 2 Substitute 20 kg for m , and v 0 for v i , K i = 1 2 ( 20 kg ) ( v 0 ) 2 = 10 ( v 0 ) 2 kg Calculation of final kinetic energy: Modify the expression in equation (2) to find the final kinetic energy of the child as follows. K f = 1 2 m ( v f ) 2 Substitute 20 kg for m , and 0 m s for v f , K f = 1 2 ( 20 kg ) ( 0 m s ) 2 = 0 kg ⋅ m 2 s 2 = 0 J ( ∵ 1 kg ⋅ m 2 s 2 = 1 J ) Calculation of initial gravitational potential energy: Modify the expression in equation (3) to find the initial gravitational potential energy of the child as follows

College Physics: A Strategic Approach (4th Edition)

4th Edition

ISBN: 9780134704180

Author: Knight

Publisher: PEARSON

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Chapter 10, Problem 77GP

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To find: The speed of the child at the bottom of the arc.

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Chapter 10, Problem 76GP

Chapter 10, Problem 78GP

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College Physics: A Strategic Approach (4th Edition)

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