Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 7 - Work and Kinetic Energy - Problems and Conceptual Exercises - Page 214: 72

Answer

a) $W=(mg)(h_{max}-h)$ b) $W=mgh$ c) $K.E_f=\frac{1}{2}mv_{\circ}^2+mgh$

Work Step by Step

(a) The required work done is given as $W=Fd$ $W=(mg)(h_{max}-h)$ (b) Work done on ball by gravity when it reaches ground level is given as $W=Fd$ $W=-(mg)(0-h)$ $W=mgh$ (c) We know that the expression for the final kinetic energy is given as $K.E_f-K.E-i=W$ $\implies K.E_f=K.E_i+W$ $\implies K.E_f=\frac{1}{2}mv_{\circ}^2+mgh$
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