College Physics (4th Edition)

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

Review & Synthesis: Chapters 6-8 - MCAT Review - Page 326: 5


The correct answer is: D. $1.5\times 10^5~\times sin~10^{\circ}$

Work Step by Step

The extra power would be required to increase the gravitational potential energy of the car. We can find the increase in gravitational potential energy each second: $\Delta U_g = mgh$ $\Delta U_g = mgd~sin~\theta$ $\Delta U_g = (1000~kg)(10~m/s^2)(15~m)~sin~10^{\circ}$ $\Delta U_g = 1.5\times 10^5~sin~10^{\circ}~J$ Since this amount of additional energy is required each second, the extra power that is required is $1.5\times 10^5~sin~10^{\circ}~W$ The correct answer is: D. $1.5\times 10^5~\times sin~10^{\circ}$
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