Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 7 - Newton's Third Law - Exercises and Problems: 39

Answer

(a) The minimum mass that will stick and not slip is 1.83 kg (b) $a = 1.32~m/s^2$

Work Step by Step

(a) To find the minimum mass, we can assume that the force of static friction is at a maximum. Then the component of the weight of block $m$ directed down the slope plus the force of static friction directed down the slope will be equal in magnitude to the weight of the 2.0-kg block. $mg~sin(\theta)+F_f = (2.0~kg)~g$ $mg~sin(\theta)+mg~cos(\theta)~\mu_s = (2.0~kg)~g$ $m~sin(\theta)+m~cos(\theta)~\mu_s = 2.0~kg$ $m = \frac{2.0~kg}{sin(\theta)+cos(\theta)~\mu_s}$ $m = \frac{2.0~kg}{sin(20^{\circ})+cos(20^{\circ})(0.80)}$ $m = 1.83~kg$ The minimum mass that will stick and not slip is 1.83 kg (b) Note that both blocks will have the same magnitude of acceleration. We can set up a force equation for the 2.0-kg block. Let $T$ be the tension in the rope. Let $M$ be the mass of this block. So; $\sum F = Ma$ $Mg-T = Ma$ $T = M(g-a)$ We can set up a force equation for the 1.83-kg block. Let $m$ be the mass of this block. $\sum F = ma$ $T - mg~sin(\theta) - F_f = ma$ $M(g-a) - mg~sin(\theta) - mg~cos(\theta)~\mu_k = ma$ $Mg - mg~sin(\theta) - mg~cos(\theta)~\mu_k = (M+m)a$ $a = \frac{Mg - mg~sin(\theta) - mg~cos(\theta)~\mu_k}{M+m}$ $a = \frac{(2.0~kg)(9.80~m/s^2) - (1.83~kg)(9.80~m/s^2)~sin(20^{\circ}) - (1.83~kg)(9.80~m/s^2)~cos(20^{\circ})(0.50)}{2.0~kg+1.83~kg}$ $a = 1.32~m/s^2$
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.