Question 33: Laws of Motion

Question

A block takes $t$ time to slide down a plane inclined at $45^{\circ}$ to the horizontal. If the surface is made smooth (frictionless), the block takes time $\frac{t}{2}$ to slide down the plane. The coefficient of friction between the block and the inclined plane is $\left(\frac{\alpha}{100}\right)$. The value of $\alpha$ is $\_\_\_\_$ .

Accepted Answer

When a block of mass $m$ is placed on a plane inclined at an angle $	heta$ :
Case 1: Rough Surface
The forces acting along the plane are the component of gravity ( $\mathrm{mg} \sin 	heta$ ) acting downwards and the kinetic friction ( $\mathrm{f}_{\mathrm{k}}=\mu \mathrm{N}=\mu \mathrm{mg} \cos 	heta$ ) acting upwards.
The net force acting along the inclined surface is $\mathrm{F}_{	ext {net }}=\mathrm{mg} \sin 	heta-\mu \mathrm{mg} \cos 	heta$
Using Newton's second law of motion $\mathrm{F}=\mathrm{ma}$, the acceleration of block on rough surface is,
$$ \mathrm{a}_{	ext {rough }}=\mathrm{g}(\sin 	heta-\mu \cos 	heta) $$
Case 2: Smooth Surface
Friction is zero ( $\mu=0$ ).
The net force acting along the inclined surface is $\mathrm{F}_{	ext {net }}=\mathrm{mg} \sin 	heta$
So, the acceleration of block on smooth surface is,
$$ \mathrm{a}_{	ext {smooth }}=\frac{\mathrm{mg} \sin 	heta}{\mathrm{~m}}=\mathrm{g} \sin 	heta $$
The block starts from rest ( $\mathrm{u}=0$ ) and slides a distance s .
Using the kinematic equation $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2:$
$$ s=\frac{1}{2} a t^2 \Rightarrow t=\sqrt{\frac{2 s}{a}} $$
For a constant distance s , the relationship is $\mathrm{t} \propto \frac{1}{\sqrt{\mathrm{a}}}$, or a $\propto \frac{1}{\mathrm{t}^2}$.
The angle of inclination is $	heta=45^{\circ}$.
The time taken on rough surface is $\mathrm{t}_{	ext {rough }}=\mathrm{t}$.
The time taken on smooth surface is $t_{	ext {smooth }}=\frac{t}{2}$.
So, the ratio of accelerations is;
$$ \frac{\mathrm{a}_{	ext {smooth }}}{\mathrm{a}_{	ext {rough }}}=\left(\frac{\mathrm{t}_{	ext {rough }}}{\mathrm{t}_{	ext {smooth }}}ight)^2=\left(\frac{\mathrm{t}}{\mathrm{t} / 2}ight)^2=(2)^2=4 $$
Substituting the acceleration formulas into the ratio :
$$ \frac{\mathrm{g} \sin 45^{\circ}}{\mathrm{g}\left(\sin 45^{\circ}-\mu \cos 45^{\circ}ight)}=4 $$
$$\Rightarrow $$ $\frac{1 / \sqrt{2}}{\left(\frac{1}{\sqrt{2}}-\mu\left(\frac{1}{\sqrt{2}}ight)ight)}=4$
$$\Rightarrow $$ $\frac{1}{1-\mu}=4$
$$\Rightarrow $$ $1=4-4 \mu$
$$\Rightarrow $$ $4 \mu=3 \Rightarrow \mu=0.75=\frac{75}{100}$
$$\Rightarrow $$ $\frac{75}{100}=\frac{\alpha}{100} \Rightarrow \alpha=75$
Therefore, the value of $\alpha$ is 75 .

Laws of Motion

Integer / Numeric Answer Type