University Physics with Modern Physics (14th Edition)

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

Chapter 7 - Potential Energy and Energy Conservation - Problems - Exercises: 7.1

Answer

A. The climber gained 661,500 J of potential energy. B. The climber lost 771,750 J (-771,750) of potential energy.

Work Step by Step

A.) We know that on day 1 the climbers starting point is 1500m of the ground, and she is climbing to 2400m. We also know she is 75kg. We need subtract her final potential energy from her initial potential energy. 1. U=mgh 2. Ui = (75)(9.8)(1500) = 1,102,500 & Uf = (75)(9.8)(2400) = 1,764,000J 3. Uf - Ui = 1,764,000 - 1,102,500 = 661,500 J We are going to use the same process as above but our initial height is 2400m and our final height is 1350m. 1. U=mgh 2. Ui = (75)(9.8)(2400) = 1,764,000 & Uf = (75)(9.8)(1350) = 992,250 3. Uf - Ui = 992,250 - 1,764,000 = -771,750
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.