Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - General Problems - Page 168: 79

Answer

(a) $W = -9.0 \times 10^4 ~J$ (b) $F = 8.2\times 10^4 ~N$ (c) $W = -2.3 \times 10^5 ~J$

Work Step by Step

(a) The work done by the snow is equal to the change in energy. We need to consider $\Delta KE$ and $\Delta PE$ starting from the moment of impact with the snow until the person came to a stop. Therefore, $W = \Delta KE + \Delta PE$ $W = -\frac{1}{2}mv^2 - mgd$ $W = \frac{1}{2}(88 ~kg)(45 ~m/s)^2 -(88 ~kg)(9.80 ~m/s^2)(1.1 ~m)$ $W = -9.0 \times 10^4 ~J$ (b) Let $F$ be the magnitude of the force. Then ($F \cdot d$) is equal to the magnitude of the work: $F\cdot d = 9.0 \times 10^4 ~J$ $F = \frac{9.0 \times 10^4 ~J}{d} = \frac{9.0 \times 10^4 ~J}{1.1 ~m}$ $F = 8.2\times 10^4 ~N$ (c) We can use work and energy to solve this question. $PE + W = KE$ $mgy - W = \frac{1}{2}mv^2 $ $W = \frac{1}{2}mv^2 - mgy$ $W = \frac{1}{2}(88 ~kg)(45 ~m/s)^2 - (88 ~kg)(9.80 ~m/s^2)(370 ~m)$ $W = -2.3 \times 10^5 ~J$
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.