University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 6 - Work and Kinetic Energy - Problems - Exercises - Page 198: 6.66

Answer

The package has a speed of 3.41 m/s after sliding 2.80 meters.

Work Step by Step

(a) $W = -F_f~d$ $W = -mg~cos(\theta)~\mu_k~d$ $W = -(5.00~kg)(9.80~m/s^2)~cos(24.0^{\circ})(0.310)(2.80~m)$ $W = -38.9~J$ (b) $W = mg~sin(\theta)~d$ $W = (5.00~kg)(9.80~m/s^2)~sin(24.0^{\circ})(2.80~m)$ $W = 55.8~J$ (c) Since the normal force acts at an angle of $90^{\circ}$ to the direction of motion, the normal force does zero work. (d) $W_{tot} = 55.8~J - 38.9~J = 16.9~J$ (e) $W_{tot} = K_2-K_1$ $W_{tot} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$ $v_2^2 = \frac{2~W_{tot}+mv_1^2}{m}$ $v_2 = \sqrt{\frac{(2)(16.9~J)+(5.00~kg)(2.20~m/s)^2}{5.00~kg}}$ $v_2 = 3.41~m/s$ The package has a speed of 3.41 m/s after sliding 2.80 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.