Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 7 - Kinetic Energy and Work - Problems: 19


25 Joules

Work Step by Step

As we know that $W_{total}=W_{gravity}+W_{w}$ Where $W_{total}$ is the total work done while $W_{gravity}$ and $W_{w}$ represent Work done by gravity and Work done by worker respectively. First of all we calculate work done by the worker $(\theta=180^{\circ} )$ because the direction of motion of block and force applied by worker are opposite $W_{w}$=$F d$$cos 180^{\circ}$ $W_{w}$=$50\times{0.5}(-1)$ $W_{w}$=$-25J$ $W_{total}=W_{gravity}+(-25)J$ As we know that total change in $K.E$ is in fact total work done so $80J$=$W_{gravity}-25J$ $W_{gravity}$=$80+25J$ $W_{gravity}$=105J If the rope had not been attached then the work done would be just due to gravity that is $105J$ and it would be $25J$ greater than current $K.E$ which is 80J.i.e $105-80=25J$
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