Question

A block starts slipping down an inclined plane and moves half a meter in half-second. How long will it take to cover the next half meter (in $sec$)?

Answer 1

Let the block of masss $m$ starts sliding down the plane witn an acceleration $a$ as shown in the figure.

FBD of the block,

Let the coefficient of friction between block and inclined surface be $\mu$
Given, initial velocity of the block of mass $m$ is $u=0$
Now we have, from the FBD
$mg\sin \theta -f=ma$..........(i)
and, $N=mg\cos \theta$
also,$f=\mu N=\mu mg\cos \theta$
Putting these values in the eq (i), we get
$a=g\sin \theta -\mu g\cos \theta$ ,which will remain constant.

For first half meter, let the time taken be $t_1$
$\Rightarrow s=u{t}_1+\frac{1}{2}a{t}_1^2$
$\frac{1}{2}=0+\frac{1}{2}a{t}_1^2$
$\Rightarrow t_1=\frac{1}{\sqrt{a}}$

For full $1m$, let time be $t_2$
$\Rightarrow s=u{t}_2+\frac{1}{2}a{t}_2^2$
$1=0+\frac{1}{2}a{t}_2^2$
$\Rightarrow t_2=\frac{\sqrt{2}}{\sqrt{a}}$

So, $\frac{t_2}{t_1}=\sqrt{2}$
$t_2=\sqrt{2}{t}_1=\sqrt{2}\times \frac{1}{2}=0.707s$
Hence, for next half meter time taken would be ($t_2-t_1)=(0.707-0.5)=0.207s$