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 = n \over k\sin \theta . The value of n is \underline\hspace2cm .

Work Power and Energy PYQ — JEE Main 2023 · 25 Jan · Shift 1 · Q65 with Solution

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Correct answer: 2

We need kinetic energy $E$ as a function of $\theta$ .

Since the object moves only horizontally, only the horizontal component of force does work.

At displacement $dx$ , work done is: $dW = F\cos\theta\, dx$

Given $F=2$ and $\theta = kx$ , $dW = 2\cos(kx)\, dx$

By work-energy theorem, $dE = dW = 2\cos(kx)\, dx$

Initially, at $x=0$ , object is at rest, so $E=0$

Thus, $E = \int_0^x 2\cos(kx)\, dx$

To avoid confusion in variable, write: $E = \int_0^x 2\cos(ks)\, ds$

Integrating, $E = 2\left[\frac{\sin(ks)}{k}\right]_0^x$

$E = \frac{2}{k}\sin(kx)$

Since $\theta = kx$ , $E = \frac{2}{k}\sin\theta$

Given: $E = \frac{n}{k}\sin\theta$

So, $n=2$

Stored correct answer = $2$

Our derived answer also gives $2$ .