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 - Problems: 14

Answer

a. $1.1 \times 10^7 J$. b. $5.0 \times 10^7 J$.

Work Step by Step

a. The thrust force and the displacement are both forward. The angle between them is 0 degrees. Use equation 6–1. From the graph we see that the displacement is 85 m. $$W=(130 \times 10^3 N)(85 m)(cos 0^{\circ}) = 1.1 \times 10^7 J$$ b. The work done by the catapult is the signed area under the force vs position graph. The shape is a trapezoid, with an area of $W = (85 m)\frac{1}{2}(1100 \times 10^3 N + 65 \times 10^3 N) = 5.0 \times 10^7 J$
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.