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 11 - Work - Exercises and Problems: 34

Answer

(a) $F = 100~N$ (b) The power output at t = 2.0 s is 400 W The power output at t = 4.0 s is 800 W The power output at t = 6.0 s is 1200 W

Work Step by Step

(a) We can find the sprinter's acceleration as: $x = \frac{1}{2}at^2$ $a = \frac{2x}{t^2}$ $a = \frac{(2)(50~m)}{(7.0~s)^2}$ $a = 2.0~m/s^2$ We then find the horizontal force $F$ acting on the sprinter. $F = ma$ $F = (50~kg)(2.0~m/s^2)$ $F = 100~N$ (b) We can find the speed at t = 2.0 s $v = a~t$ $v = (2.0~m/s^2)(2.0~s)$ $v = 4.0~m/s$ We then find the power at t = 2.0 s $P = F~v$ $P = (100~N)(4.0~m/s^)$ $P = 400~W$ The power output at t = 2.0 s is 400 W. Next, we find the speed at t = 4.0 s: $v = a~t$ $v = (2.0~m/s^2)(4.0~s)$ $v = 8.0~m/s$ We can find the power at t = 4.0 s: $P = F~v$ $P = (100~N)(8.0~m/s^)$ $P = 800~W$ The power output at t = 4.0 s is 800 W. We can find the speed at t = 6.0 s $v = a~t$ $v = (2.0~m/s^2)(6.0~s)$ $v = 12.0~m/s$ We can find the power at t = 6.0 s $P = F~v$ $P = (100~N)(12.0~m/s^)$ $P = 1200~W$ The power output at t = 6.0 s is 1200 W.
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.