A disc of mass $100 \mathrm{~g}$ slides down from rest on an inclined plane of $30^{\circ}$ and come to rest…
- 0.106 J
- 0.05 J
- 0.306 J
- 0.2 J
Solution
$ v^2=u^2-2 a s $ or $ u=\sqrt{2 a s} $ $(\because v=0)$ $\because$ Acceleration of a block on a horizontal, $ \begin{gathered} a=\mu g \\ \left.\Rightarrow \quad a=0.2 \times 10=2 \mathrm{~m} / \mathrm{s}^2 \quad \text { (Given, } \mu=0.2\right) \end{gathered} $ After putting the value of $a$ in Eq. (i), we get $ \begin{array}{rlrl} \therefore & u & =\sqrt{2 \times 2 \times 1} \\ & =2 \mathrm{~m} / \mathrm{s} \end{array} \quad(\because s=1 \mathrm{~m}, \text { given }) $ From above diagram, the frictional force applied on the disc inclined plane $r$ is $ \begin{aligned} F & =\mu N \\ & =\mu m g \cos 30^{\circ} \end{aligned} $ and the net acceleration force down the inclined plane, $ m g \sin 30^{\circ}-f=m g \sin 30^{\circ}-\mu m g \cos 30^{\circ}=m a_1 $
Hence, from the third equation of motion, $ v_1^2=u_1^2+2 a_1 s $ When $s$ is distance travelled by the disc, $ \begin{array}{rlrl} & v_1=u=2 \mathrm{~m} / \mathrm{s}, u_1=0 \\ \therefore & (2)^2 & =2 a_1 s \text { or } s=\frac{2}{a_1} \end{array} $ Putting the value of $a_1$ from Eqs. (i), we get $ \begin{aligned} \Rightarrow \quad s & =\frac{2}{5(1-\sqrt{3} \mu)} \\ & =\frac{2}{5(1-\sqrt{3} \times 0.2)}=0.612 \mathrm{~m} \end{aligned} $ Hence, the work done by the frictional force, $W=$ work done on inclined plane + work done on horizontal plane. $ =\left(\mu m g \cos 30^{\circ}\right) s+\frac{1}{2} m u^2 $ Putting the given values, we get $ \begin{aligned} & =0.2 \times \frac{100}{1000} \times 10 \times \cos 30^{\circ} \times 0.612+\frac{1}{2} \times \frac{100}{1000} \times(2)^2 \\ & =0.306 \mathrm{~J} \end{aligned} $
Asked in: AP EAMCET 2019 (20 Apr Shift 2)