Explanation Draw the digram illustrating the forces on the block. At the unstretched position of the spring consider potential energy to be zero for block of mass 30.0 kg and for the 20.0 kg block consider it to be at lowest point just before the system is released from rest. Write the expression for kinetic energy of the system. K = 1 2 m v 2 (I) Here, K is kinetic energy, v is velocity and m is mass. Write the expression for gravitational potential energy as. U g = m g h (II) Here, U g is gravitational potential energy, m is mass of the block, g is acceleration due to gravity and h is height. Write the expression for elastic energy stored in the spring as. U s = 1 2 k x 2 (III) Here, U s is elastic energy, k is force constant and x is compression in the spring. Total energy of the system remains conserved therefore total initial energy of the system is equal to total final energy for both the masses. As the system is released from rest its initial kinetic energy is zero and final kinetic energy is due to both the masses as both will move together. Write the expression for kinetic energy at final position. K f = 1 2 ( m 1 + m 2 ) v 2 Here, m 1 is the mass of block placed on the inclined and m 2 is the mass of the block attached to the spring. Write the expression for conservation of energy for this system. U g i + U s + K i = K f + U g f (IV) Substitute m 2 g h for U g i , 0 for K i , 1 2 k x 2 for U s , 1 2 ( m 1 + m 2 ) v 2 for K f , m 1 g h sin θ for U g f and h for x because stretch of the spring is equal to the height moved by the block in equation (IV)

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term

9th Edition

ISBN: 9781305932302

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Question

Chapter 8, Problem 64AP

To determine

The speed of each block when the spring is again unstretched.

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Chapter 8, Problem 63AP

Chapter 8, Problem 65AP

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