#### 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 $=0.1178\times1.164+0.577=0.714$