College Physics (4th Edition)

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

Chapter 9 - Problems - Page 360: 9

Answer

The small piston must be pushed down a distance of $1.0~meter$

Work Step by Step

Let $F_A$ be the force exerted by the large piston on the car. The work that this piston does on the car is $F_A~d$, where $d$ is the distance that the force is exerted. In order to lift the car, the work done on the small piston must be equal to the work done by the large piston. We can find the distance $d'$ that the small piston must be pushed down with a force of $F_a$: $F_a~d' = F_A~d$ $d' = \frac{F_A~d}{F_a}$ By Pascal's principle, $\frac{F_A}{F_a} = \frac{A}{a}$: $d' = \frac{F_A~d}{F_a}$ $d' = \frac{A~d}{a}$ $d' = 100.0~d$ $d' = (100.0)~(1.0~cm)$ $d' = 1.0~m$ The small piston must be pushed down a distance of $1.0~meter$
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