Question

A body of mass $m$ resting on a smooth horizontal plane starts moving under a constant force $F$. during its rectilinear motion, the angle $\theta$ between the direction of force and horizontal plane varies as $\theta =kx$, 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?

• A. $\sqrt{\frac{F\sin \theta }{mk}}$
• B. $\sqrt{\frac{2F\sin \theta }{mk}}$
• C. $\sqrt{\frac{F\cos \theta }{mk}}$
• D. $\sqrt{\frac{2F\cos \theta }{mk}}$

Answer 1

Answer: (B)

$\sqrt{\frac{2F\sin \theta }{mk}}$

At any $x$.

Writing $NLM$ equasia.

$\begin{array}{c}F\cos \theta =ma\theta =kx\\ \Rightarrow F\cos kx=ma\\ \Rightarrow F\cos kx=m.v\frac{dv}{dx}\\ \Rightarrow m{\int }_0^{v}vdv=F{\int }_0^x\cos kxdx\\ \Rightarrow \frac{m{v}^2}{2}=\frac{F\sin kx}{k}\\ \Rightarrow v^2=\frac{2F\sin \theta }{km}\\ \Rightarrow v=\sqrt{\frac{2F\sin \theta }{km}}\end{array}$

$\therefore$ Option $B$ is correct.