## Physics (10th Edition)

Published by Wiley

# Chapter 6 - Work and Energy - Problems - Page 170: 70

#### Answer

$\theta=2^o$

#### Work Step by Step

We call the force of the engine when it goes up $F_u$ and when it goes down $F_d$. We have $$P_{up}-P_{down}=47hp=3.5\times10^4W$$ $$(F_{u}-F_{d})v=3.5\times10^4W$$ We know $v=27m/s$, so $$F_{u}-F_{d}=1.3\times10^3N (1)$$ - When the car goes up: We call the force of the engine when it goes up $F_u$ and the combined force of air resistance and friction $R$, both of which influence the car's motion. $mg\sin\theta$ influences the car's motion, too, by opposing it. The car goes at a constant velocity, so $\sum F=0$ $$F_u-R-mg\sin\theta=0(2)$$ - When the car goes down: We call the force of the engine when it goes down $F_d$, which propels the motion. $mg\sin\theta$, this time, supports the car's downward motion. $R$ still opposes it. Similarly, here $\sum F=0$ $$F_d+mg\sin\theta-R=0 (3)$$ Subtract (3) from (2): $$F_u-F_d-2mg\sin\theta=0$$ We know $F_{u}-F_{d}=1.3\times10^3N$ and $m=1900kg$, so $$\sin\theta=0.035$$ $$\theta=2^o$$

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