An object of mass '$m$' initially at rest on a smooth horizontal plane starts moving under the action of force $\mathrm{F}=2 \mathrm{~N}$. In the process of its linear motion, the angle $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta=\mathrm{kx}$, 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 $\mathrm{E}=\frac{\mathrm{n}}{\mathrm{k}} \sin \theta$. The value of $n$ is $\_\_\_\_$.

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Work, Power & Energy

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

θ

(as shown in figure) between the direction of force and horizontal varies as

θ = 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 θ

. The value of

n

____

Answer: 2

The object accelerates under the horizontal part of the force,

F cos θ = 2 cos (k x)

. By the work-energy theorem the kinetic energy equals the work done from the start,

E = \int_{0}^{x} F cos (k x) d x = 2 \int_{0}^{x} cos (k x) d x = \frac{2}{k} sin (k x)

. Since

θ = k x

, this is

E = \frac{2}{k} sin θ

, and comparison with

\frac{n}{k} sin θ

gives

n = 2

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Integer answer