JEE Physics 2023 Question with Solution

Given: The object starts from rest on a smooth horizontal plane. The applied force has magnitude $F = 2 \, \text{N}$ and the angle with the horizontal varies as $\theta = kx$.

Find: The value of $n$ in $E = \frac{n}{k} \sin \theta$.

Using the work-energy theorem:

$$W = \Delta K.E = E$$

Only the horizontal component of force does work, so

$$E = \int_0^x F \cos \theta \, dx$$

Substituting $F = 2$ and $\theta = kx$,

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

Evaluating the integral,

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

Since $\theta = kx$,

$$E = \frac{2}{k} \sin \theta$$

Comparing with $E = \frac{n}{k} \sin \theta$, we get $n = 2$.

Therefore, the required value is $2$.

Given: $F = 2 \, \text{N}$ and $\theta = kx$.

Find: The coefficient $n$.

Since displacement is horizontal, directly take the working component of force as $F \cos \theta = 2\cos(kx)$. Then kinetic energy gained equals work done:

$$E = \int_0^x 2\cos(kx) \, dx = \frac{2}{k}\sin(kx) = \frac{2}{k}\sin\theta$$

Hence, comparing with $E = \frac{n}{k} \sin \theta$, the correct value is $n = 2$.