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 - Page 305: 49

Answer

The length of the ramp should be 120 meters.

Work Step by Step

We can find the initial kinetic energy of the truck. $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(15,000~kg)(35~m/s)^2$ $KE = 9.19\times 10^6~J$ Let $d$ be the distance the truck travels along the ramp. We can find an expression for the height $h$. $\frac{h}{d} = sin(\theta)$ $h = d~sin(\theta)$ The final potential energy will be equal to the sum of the initial kinetic energy and the work done by friction. $PE = KE+W_f$ $PE - W_f = KE$ $mgh - (-mg~\mu_k~d) = KE$ $mgd~sin(\theta) +mg~\mu_k~d = KE$ $d = \frac{9.19\times 10^6~J}{(15,000~kg)(9.80~m/s^2)~[sin(6.0^{\circ}) +0.40]}$ $d = 120~m$ The length of the ramp should be 120 meters.
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.