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: 5


$F_{max} = 1500~N$

Work Step by Step

The impulse is equal to the area under the force versus time graph. We can find the area under the graph. $Area = \frac{1}{2}(F_{max})(8\times 10^{-3}~s)$ $Area = (F_{max})(4\times 10^{-3}~s)$ To find the value of $F_{max}$, we can equate the area under the graph to the impulse. $(F_{max})(4\times 10^{-3}~s) = 6.0~N~s$ $F_{max}= \frac{6.0~N~s}{(4\times 10^{-3}~s)}$ $F_{max} = 1500~N$
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