Question

At $t=0$, a force $F=kt$ is applied on a block making an angle $\alpha$ with the horizontal. Assume surfaces to be smooth. Find the velocity of the body at the time of breaking off the plane :

• A. $\frac{mg\cos \alpha }{2k\sin \alpha }$
• B. $\frac{m^2{g}^2\cos \alpha }{2k{\sin }^2\alpha }$
• C. $\frac{m{g}^2\cos \alpha }{2k{\sin }^2\alpha }$
• D. $\frac{m{g}^2\cos \alpha }{2k\sin \alpha }$

Answer 1

Answer: (C)

$\frac{m{g}^2\cos \alpha }{2k{\sin }^2\alpha }$

From FBD, $F\sin \alpha =mg$
$kt\sin \alpha =mg$ as$(F=kt)$ (Since for take off N=0)
$t=\frac{mg}{k\sin \alpha }......\left(1\right)$
now, $F\cos \alpha =ma\Rightarrow kt\cos \alpha =m\frac{dv}{dt}$
$dv=\frac{k}{m}\cos \alpha tdt$
or,${\int }_0^{v}dv=\frac{k}{m}\cos \alpha {\int }_0^{t}tdt$
or,$v=\frac{k}{2m}\cos \alpha t^2$
using (1), the velocity at $t$ is $v=\frac{k}{2m}\cos \alpha \times \frac{m^2{g}^2}{k^2{\sin }^2\alpha }=\frac{m{g}^2\cos \alpha }{2k{\sin }^2\alpha }$