College Physics (4th Edition)

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

Chapter 6 - Problems - Page 232: 84

Answer

(a) The engine's average power output is $3.64\times 10^5~Watts$ (b) The volume of gasoline that is consumed is 79 mL

Work Step by Step

(a) We can find the car's change in kinetic energy: $KE = \frac{1}{2}mv^2 = \frac{1}{2}(1000.0~kg)(40.0~m/s)^2 = 8.00\times 10^5~J$ This energy is only 22% of the energy released by burning gasoline. We can find the total energy released: $0.22~Energy = 8.00\times 10^5~J$ $Energy = \frac{8.00\times 10^5~J}{0.22}$ $Energy = 3.64\times 10^6~J$ This much energy is released in a time of 10.0 seconds. We can find the power output: $Power = \frac{Energy}{Time} = \frac{3.64\times 10^6~J}{10.0~s} = 3.64\times 10^5~Watts$ The engine's average power output is $3.64\times 10^5~Watts$ (b) We can find the volume of gasoline that is consumed: $V = \frac{3.64\times 10^6~J}{46\times 10^6~J/L} = 0.079~L = 79~mL$ The volume of gasoline that is consumed is 79 mL.
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