Question

A block begins to slide down a rough inclined plane of angle ${45}^{\circ }$ and moves $1m$ in $\sqrt[4]{42}\text{second}$. What is the coefficient of friction between the plane and the block? $g=10m/s^2$

• A. $0.4$
• B. $0.5$
• C. $0.6$
• D. $0.8$

Answer 1

The correct option is D $0.8$

The said condition can be shown as


From the figure we have
$N=mg\cos {45}^{\circ }$
$f=\mu N=\mu mg\cos {45}^{\circ }$
So, from force equation we have,
$mg\sin {45}^{\circ }-f=ma$
$\Rightarrow mg\sin {45}^{\circ }-\mu mg\cos {45}^{\circ }=ma$
$\Rightarrow a=g(\sin {45}^{\circ }-\mu \cos {45}^{\circ })=10(\frac{1}{\sqrt{2}}-\frac{\mu }{\sqrt{2}})m/s^2$

Given, total length of the incline $s=1\text{m}$ and time taken, $t=\sqrt[4]{42}\text{sec}$
$\therefore s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow 1=0+\frac{1}{2}\times 10(\frac{1}{\sqrt{2}}-\frac{\mu }{\sqrt{2}})(\sqrt[4]{42}{)}^2$
$\Rightarrow \mu =0.8$