College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 14 - Problems - Page 533: 8


The descending mass that should be used is m=2.72 kg

Work Step by Step

The potential energy of the descending mass is converted to heat inside the water (usually use Potential energy if there is a height). h= (30)(1.25m) As it descends 30 times m= ? $\Delta$P = 1000 J g= 9.8 $m/s^{2}$ Let $\Delta$$P_{grav}$ = $\Delta$$E_{interal energy}$ $\Delta$P=mgh Rearange the equation to make m your subject of the equation m=$\frac{\Delta U}{g.h}$ =$\frac{1000J}{(9.8 m/s^{2})(30)(1.25m))}$ =2.72kg
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