## College Physics (7th Edition)

Published by Pearson

# Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 140: 88

#### Answer

a). The free body diagram is as shown. b). $F_{f}-mgsin\theta=ma$ or, $F_{f}$=Friction coefficient $\times N=m(a+gsin\theta)$ Again, $N-mgcos\theta=0$ or, $N=mgcos\theta$ c). Friction coefficient $=0.714$

#### Work Step by Step

a). The free body diagram is as shown. b). $F_{f}-mgsin\theta=ma$ or, $F_{f}$=Friction coefficient $\times N=m(a+gsin\theta)$ Again, $N-mgcos\theta=0$ or, $N=mgcos\theta$ c). From above 2 equations, we have friction coefficient =$\frac{m(a+gsin\theta)}{mgcos\theta}=\frac{a}{gcos\theta}+tan\theta=\frac{a}{9.8cos30^{\circ}}+tan30^{\circ}=0.1178a+0.577$ Now, $a=\frac{v^{2}-u^{2}}{2S}=\frac{1.6^{2}}{2\times1.1}=1.164m/s^{2}$ Putting this value of a in previous equation, Friction coefficient $=0.1178\times1.164+0.577=0.714$

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