Q 11-05-098JEE MainJEE Main 2023 (25 Jan, Shift 1)Medium
An object of mass $m$ initially at rest on a smooth horizontal plane starts moving under the action of force $F=2\ \text{N}$. In the process of its linear motion, the angle $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta=kx$, 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 $E=\dfrac nk\sin\theta$. The value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 2
By the work–energy theorem, only the horizontal component does work:
$$E=\int_0^xF\cos(kx)\,dx=\frac{2}{k}\sin(kx)=\frac2k\sin\theta$$
So $n=2$.
Solution by Sreeraj P, M.Sc Physics