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A body of mass m resting on a smooth horizontal plane starts moving under a constant force F. During its rectilinear motion, the angle $θ$ between the direction of force and horizontal plane varies as $θ = k x$ ,where k is a constant and x is the distance travelled by the body from the initial position. What is the velocity of the body

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\sqrt{\frac{F \sin θ}{mk}}

\sqrt{\frac{2F \sin θ}{mk}}

\sqrt{\frac{F \cos θ}{mk}}

\sqrt{\frac{2F \cos θ}{mk}}

answer is B.

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Detailed Solution

a = \frac{F \cos k x}{m}

$\frac{dv}{dt} = \frac{F \cos k x}{m}$

$\frac{dv}{d x} . \frac{d x}{dt} = \frac{F \cos k x}{m}$

$\int_{0}^{v} vdv = \int_{0}^{x} \frac{F \cos k x}{m} d x$

$\frac{v^{2}}{2} = \frac{F \sin k x}{mk}$

$v^{2} = \frac{2F \sin k x}{mk} = \sqrt{\frac{2F \sin k x}{mk}} = \sqrt{\frac{2F \sin θ}{mk}}$

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