Question

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

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

Answer 1

The correct option is D $0.8$

Let $a$ be the acceleration of the block down the inclined plane. The given question can be depicted as shown below

The coefficient of friction between the block and the incline plane is $\mu$
Now, from the FBD we have
$mg\sin {45}^{\circ }-f=ma$......(1)
and, $N=mg\cos {45}^{\circ }$
Friction force, $f=\mu N=\mu mg\cos {45}^{\circ }$
On putting the value of $f$ in eq (1) we get
$mg\sin {45}^{\circ }-\mu mg\cos {45}^{\circ }=ma$
$\Rightarrow a=g(\sin {45}^{\circ }-\mu \cos {45}^{\circ })=\frac{10(1-\mu )}{\sqrt{2}}m/s^2$

Using the equation of motion
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow 1=0+\frac{1}{2}\times \frac{10(1-\mu )}{\sqrt{2}}\times {\left(\sqrt[4]{42}\right)}^2$$\Rightarrow \mu =0.8$