Question

At the moment $t=0$ the force $F=at$ is applied to a small body of mass $m$ resting on a smooth horizontal plane ($a$ is a constant). The permanent direction of this force forms an angle $\alpha$ with the horizontal as shown in the figure above.

Find the distance traversed by the body up to this moment :

• A. $s=\frac{m^2{g}^3\cos \alpha }{6a^2{\sin }^3\alpha }$
• B. $s=\frac{2m^2{g}^3\cos \alpha }{6a^2{\sin }^3\alpha }$
• C. $s=\frac{m{g}^3\cos \alpha }{6a^2{\sin }^3\alpha }$
• D. $s=\frac{2m{g}^3\cos \alpha }{6a^2{\sin }^3\alpha }$

Answer 1

The correct option is A $s=\frac{m^2{g}^3\cos \alpha }{6a^2{\sin }^3\alpha }$

Initially the forces acting on the block are:

$F^{\prime }=mg+F+N$
where, $N$ is normal reaction,

$m$ is mass of block,

$g$ is acceleration due to gravity,

Given that:

At $t=0\Rightarrow F^{\prime }=ma$ and $F=at$
$\Rightarrow ma=mg+at+N$

Force along X-axis, $ma=F\cos \alpha$.............(1)
Force along Y-axis, $mg=N+F\sin \alpha$ .................(2)
For $N=0$, Force along Y-axis, $mg=F\sin \alpha$
$\Rightarrow mg=at\sin \alpha$
$\Rightarrow t=\frac{mg}{a\sin \alpha }$...............(3)

Also, from (1) we can write
$\Rightarrow m\left(\frac{dv}{dt}\right)=F\cos \alpha$ ...................(since, $a=\frac{dv}{dt}$)
$\Rightarrow mdv=at\cos \alpha dt$

Integrating with respect to $t$, we get

$m{\int }_0^{v}dv=a\cos \alpha {\int }_0^{t}tdt$

$mv=a\cos \alpha \frac{t^2}{2}$

$v=\frac{a\cos \alpha t^2}{2m}$

$v=\frac{a\cos \alpha (\frac{mg}{a\sin \alpha }{)}^2}{2m}$ .........from(3)

$v=\frac{m{g}^2\cos \alpha }{2a{\sin }^2\alpha }$

We know, the distance traversed by the body up to this moment is given by
$ds=vdt$
$ds=\frac{a\cos \alpha t^2}{2m}dt$

Integrating with respect to t, we get
$s=\frac{a\cos \alpha }{2m}{\int }_0^t{t}^{2}dt$

$s=\frac{a\cos \alpha }{2m}\frac{t^3}{3}$

$s=\frac{a\cos \alpha }{2m}\frac{(\frac{mg}{a\sin \alpha }{)}^3}{3}$ .......................from(3)

$s=\frac{m^2{g}^3\cos \alpha }{6a^2{\sin }^3\alpha }$