Question

An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force $F=2N$. In the process of its linear motion, the angle $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta =kx$, 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

Answer: 2

$2\cos (kx)=\frac{mvdv}{dx}$
$\underset{0}{\overset{v}{\int }}vdv=2\underset{0}{\overset{x}{\int }}\cos (kx)dx$
$\frac{m{v}^2}{2}=\frac{2}{k}\sin kx$
$K.E=\frac{2}{k}\sin \theta$
$n=2$