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


$F_{max} = 1000~N$

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. $(F_{max})(6\times 10^{-3}~s) = 6.0~N~s$ $F_{max}= \frac{6.0~N~s}{(6\times 10^{-3}~s)}$ $F_{max} = 1000~N$
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.