Question

An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force $F=2 N$. In the process of its linear motion, the angle $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta=k x$, where $k$ is a constant and $x$ is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be $E=\frac{n}{k} \sin \theta$. The valte of $n$ is

Answer 1

$ 2 \cos ( kx )=\frac{ mvd v}{ dx } $
$ \int\limits_0^{ v } vdv =2 \int\limits_0^{ x } \cos ( kx ) dx $
$ \frac{ m v^2}{2}=\frac{2}{ k } \sin kx $
$ K.E =\frac{2}{ k } \sin \theta $
$n = 2$