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4-2.Friction
hard
A block of mass $m$ (initially at rest) is sliding up (in vertical direction) against a rough vertical wall with the help of a force $F$ whose magnitude is constant but direction is changing. $\theta = {\theta _0}t$ where $t$ is time in sec. At $t$ = $0$ , the force is in vertical upward direction and then as time passes its direction is getting along normal, i.e., $\theta = \frac{\pi }{2}$ .The value of $F$ so that the block comes to rest when $\theta = \frac{\pi }{2}$ , is 
A$\frac{{mg \times \pi }}{{2{\theta _o}}}$
B$\frac{{mg \times \pi }}{{2\left( {1 - \mu } \right){\theta _o}}}$
C$\frac{{mg \times \pi }}{{\left( {1 - \mu } \right)}}$
D$\frac{{mg \times \pi }}{{2\left( {1 - \mu } \right)}}$
Solution
$F_{\text {net }}=m a=F \cos \theta-m g-\mu F \sin \theta$
$\int_{0}^{0} \mathrm{mdv}=\int_{0}^{\mathrm{t}_{0}}(\mathrm{F} \cos \theta-\mathrm{mg}-\mu \mathrm{F} \sin \theta) \mathrm{dt}$
$\frac{\pi}{2}=\theta_{0} \times t_{0} \& \theta=\theta_{0} \times t$
$\Rightarrow F=\frac{m g \times \pi}{2(1-\mu)}$
$\int_{0}^{0} \mathrm{mdv}=\int_{0}^{\mathrm{t}_{0}}(\mathrm{F} \cos \theta-\mathrm{mg}-\mu \mathrm{F} \sin \theta) \mathrm{dt}$
$\frac{\pi}{2}=\theta_{0} \times t_{0} \& \theta=\theta_{0} \times t$
$\Rightarrow F=\frac{m g \times \pi}{2(1-\mu)}$
Standard 11
Physics
