Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 206: 45c



Work Step by Step

Here we’ll use the equation for the change in thermal energy, which we had previously found: $\Delta E_{th} = f_{k}d = μ_{k}F_{N}d$ Solving for $μ_{k}$, we get: $μ_{k} = \frac{\Delta E_{th}}{F_{N}d}$ In Problem 45c, we found that $\Delta E_{th}$ = 30.1 J, and we also know that d = 4.06 m. So all we need to find is the normal force with Newton’s second law for vertical components. Since the block is not moving vertically, $F_{net, y}$ = 0. The forces acting vertically are the gravitational force, the inclined applied force and the normal force, so: 0 = $F_{g} + F_{app,y} + F_{N}$ 0 = $–mg + F sin\theta + F_{N}$ $F_{N} = mg – F_{app} sin\theta$ Plugging all this in we get: $μ_{k} = \frac{\Delta E_{th}}{F_{N}d}$ $μ_{k} = \frac{\Delta E_{th}}{(mg – F_{app} sin\theta)d}$ $μ_{k} = \frac{30.1 J}{((3.57 kg)(9.8 m/s²) – (7.68 N) sin 15^{\circ})(4.06 m)}$ = 0.225
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.