Explanation Given data: Refer to the equation (P10.48) in the textbook for the linear variation of the restoring force as a function of distance below equilibrium of the trampoline. Mass of the gymnast ( m ) is 72 kg. The gymnast jumps on the trampoline. At the lowest point of his motion, he is 0.80 m below equilibrium. As the given system is assumed as a spring system, the stretching distance of the spring ( x ) is 0.80 m. Formula used: Refer to the (10.16) in the textbook, and write the expression for stored elastic potential energy as follows. U s = 1 2 k x 2 (1) Here, k is the spring constant, and x is the stretching length of the spring. Refer to the equation (10.14) in the textbook, and write the expression for final gravitational potential energy of the for the gymnast as follows. U g = m g y (2) Here, g is the acceleration due to the gravity, which is 9.8 m s 2 , and y is the height at which the gymnast reaches from his lowest point of motion, which is to be determined. Explanation: The slope of the graph in given figure gives the spring constant. At 0.50 m of the distance below equilibrium, the restoring force is 2,000 N. F = 2 , 000 N x = 0.50 m Write the expression for the spring constant as follows. k = F x Substitute 2,000 N for F , and 0.50 m for x , k = 2 , 000 N 0.50 m = 4 , 000 N m Calculation of amount of elastic potential energy stored when the trampoline stretched below equilibrium by 0.80 m ( U s ) : Substitute 4 , 000 N m for k , and 0.80 m for x in equation (1), U s = ( 1 2 ) ( 4 , 000 N m ) ( 0.80 m ) 2 = ( 1 2 ) ( 4 , 000 N m ) ( 0

College Physics: A Strategic Approach (4th Edition)

4th Edition

ISBN: 9780134704180

Author: Knight

Publisher: PEARSON

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Chapter 10, Problem 48P

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To find: The height rose by the gymnast from the lowest point of his motion.

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Chapter 10, Problem 47P

Chapter 10, Problem 49P

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College Physics: A Strategic Approach (4th Edition)

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