College Physics (4th Edition)

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

Chapter 6 - Problems - Page 228: 27


It will take Brad 11.1 hours to lose 0.50 kg of fat.

Work Step by Step

We can find the change in gravitational potential energy when the barbell is lifted to a height of $2.0~m$: $U_g = mgh = (50.0~kg)(9.80~m/s^2)(2.0~m) = 980~J$ Each lift of the barbell requires $980~J$ of work. Therefore, Brad does $2940~J$ of work each minute since he lifts the barbell three times each minute. This work is only 10% of the energy that Brad gets from burning fat. Therefore, Brad's body uses $29,400~J$ of energy each minute. We can find the total energy from burning $0.50~kg$ of fat: $(500~grams)(39,000~J/gram) = 1.95\times 10^7~J$ We can find the time required to use this much energy: $t = \frac{1.95\times 10^7~J}{29,400~J/min} = 663.3~minutes = 11.1~hours$ It will take Brad 11.1 hours to lose 0.50 kg of fat.
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