## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 6 - Problems - Page 233: 99

#### Answer

(a) The average power is $94.0~Watts$ (b) The chemical energy that is used in this hike is $2.03\times 10^6~J$ (c) A person would use 485 kilocalories of food energy.

#### Work Step by Step

(a) We can find the increase in gravitational potential energy: $mgh = (70~kg)(9.80~m/s^2)(740~m) = 507,640~J$ We can find the average power output: $Power = \frac{Energy}{Time} = \frac{507,640~J}{(1.5)(3600~s)} = 94.0~Watts$ The average power is $94.0~Watts$ (b) The mechanical energy output is 25% of the chemical energy that is used in this hike. We can find the chemical energy $E_c$ that is used: $0.25E_c = 507,640~J$ $E_c = \frac{507,640~J}{0.25}$ $E_c = 2.03\times 10^6~J$ The chemical energy that is used in this hike is $2.03\times 10^6~J$ (c) We can convert the chemical energy to units of kilocalories: $E_c = 2.03\times 10^6~J \times \frac{1~kcal}{4184~J} = 485~kcal$ A person would use 485 kilocalories of food energy.

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