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 239: 4

Answer

$F_{max} = 1500~N$

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 as: $Area = \frac{1}{2}(F_{max})(8\times 10^{-3}~s)$ $Area = (F_{max})(4\times 10^{-3}~s)$ To find the value of $F_{max}$, we can equate the area under the graph to the impulse. $(F_{max})(4\times 10^{-3}~s) = 6.0~N~s$ $F_{max}= \frac{6.0~N~s}{(4\times 10^{-3}~s)}$ $F_{max} = 1500~N$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.