Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 11 - Impulse and Momentum - Exercises and Problems - Page 287: 5

Answer

$F_{max} = 1000~N$

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 as: $Area = \frac{1}{2}(F_{max})(2\times 10^{-3}~s)+(F_{max})(4\times 10^{-3}~s)+\frac{1}{2}(F_{max})(2\times 10^{-3}~s)$ $Area = (F_{max})(6\times 10^{-3}~s)$ To find the value of $F_{max}$, we can equate the area under the graph to the impulse. $(F_{max})(6\times 10^{-3}~s) = 6.0~N~s$ $F_{max}= \frac{6.0~N~s}{(6\times 10^{-3}~s)}$ $F_{max} = 1000~N$
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