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 10 - Interactions and Potential Energy - Exercises and Problems: 51

Answer

(a) $v = 1.7~m/s$ (b) If we round off, the package barely makes it onto the truck with a final kinetic energy of zero.

Work Step by Step

(a) The sum of the kinetic energy and the potential energy in the truck will be equal to the initial energy stored in the spring. $KE+PE=U_s$ $KE = U_s-PE$ $\frac{1}{2}mv^2 = \frac{1}{2}kx^2-mgh$ $v^2 = \frac{kx^2-2mgh}{m}$ $v = \sqrt{\frac{kx^2-2mgh}{m}}$ $v = \sqrt{\frac{(500~N/m)(0.30~m)^2-(2)(2.0~kg)(9.80~m/s^2)(1.0~m)}{2.0~kg}}$ $v = 1.7~m/s$ (b) We can find the work that the sticky spot does on the package. $W_f = -mg~\mu_k~d$ $W_f = -(2.0~kg)(9.80~m/s^2)(0.30)(0.50~m)$ $W_f = -2.94~J$ Let's assume that the final kinetic energy is zero. We can find the maximum possible height of the package. $PE = U_s+W_f$ $mgh = \frac{1}{2}kx^2+W_f$ $h = \frac{\frac{1}{2}kx^2+W_f}{mg}$ $h = \frac{\frac{1}{2}(500~N/m)(0.30~m)^2-2.94~J}{(2.0~kg)(9.80~m/s^2)}$ $h = 0.998~m \approx 1.0~m$ If we round off, the package barely makes it onto the truck with a final kinetic energy of zero.
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.