A block of mass m is placed on a smooth horizontal surface. A for… - Vidyadip

MCQ

A block of mass $m$ is placed on a smooth horizontal surface. A force making an angle $\theta$ with the horizontal starts acting on the block. The magnitude of the force is constant but its direction with the horizontal changes as $\theta = a+ bs$, where $a$ and $b$ are constants and $s$ is the distance covered by the block. If $|F| = 2mb$, find the velocity of the block as function of the angle $\theta$ .

A
$v = 4{(\cos \theta \, + \,\cos a)^{\frac{1}{2}}}$
✓
$v = 2{(\sin \theta \, - \sin a)^{\frac{1}{2}}}$
C
$v = 4{(\sin \theta \, - \sin a)^{\frac{1}{2}}}$
D
$v = 2{(\cos \theta \, + \,\cos a)^{\frac{1}{2}}}$

✓

Answer

Correct option: B.

$v = 2{(\sin \theta \, - \sin a)^{\frac{1}{2}}}$

b
By Newton's second law of motion.

$\mathrm{F} \cos \theta=\mathrm{ma}$

$\Rightarrow 2 \mathrm{mb} \cos (\mathrm{a}+\mathrm{bs})=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$

$\Rightarrow \frac{d v}{d t}=2 b \cos (a+b s)$

$\Rightarrow v \frac{d v}{d s}=2 b \cos (a+b s)$

$\Rightarrow \int_{0}^{v} v d v=2 b \int_{0}^{s} \cos (a+b s) d s$

$\Rightarrow\left[\frac{v^{2}}{2}\right]_{0}^{v}=\left[\frac{2 b \sin (a+b s)}{b}\right]_{0}^{s}$

$\Rightarrow v^{2}=u[\sin (a+b s)-\sin a]$

$\Rightarrow \mathrm{v}=2(\sin \theta-\sin \mathrm{a})^{1 / 2}(\because \theta=\mathrm{a}+\mathrm{bs})$

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