Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 9 - Impulse and Momentum - Exercises and Problems - Page 240: 8

Answer

After the force ends, the object's speed is 2.0 m/s in the original direction.

Work Step by Step

We can use the graph to find the impulse exerted on the object. The impulse is equal to the area under the force versus time graph. $J = F~t$ $J = (2~N)(1.0~s)$ $J = 2.0~N~s$ We then use the impulse to find the final momentum $p_f$. $p_f = p_0+J$ $p_f = m~v_0+J$ $p_f = (2.0~kg)(1.0~m/s)+2.0~N~s$ $p_f = 4.0~N~s$ We then find the velocity $v_f$ after the force ends; $m~v_f = p_f$ $v_f = \frac{p_f}{m}$ $v_f = \frac{4.0~N~s}{2.0~kg}$ $v_f = 2.0~m/s$ After the force ends, the object's speed is 2.0 m/s in the original direction.
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