JEE Physics 2023 Question with Solution

Given: mass of the body is $1 \, \text{kg}$ and the force varies with time.

Find: the power developed by the force at $t = 2 \, \text{s}$.

From the solution working, the force used is

$$\vec{F} = t \, \hat{i} + 3t^2 \, \hat{j}$$

Since $m = 1 \, \text{kg}$, the acceleration is

$$\vec{a} = \frac{\vec{F}}{m} = \vec{F}$$

Therefore,

$$\vec{a} = t \, \hat{i} + 3t^2 \, \hat{j}$$

Velocity is obtained by integrating acceleration:

$$\vec{v} = \int \vec{a} \, dt = \int (t \, \hat{i} + 3t^2 \, \hat{j}) \, dt = \frac{t^2}{2} \, \hat{i} + t^3 \, \hat{j}$$

At $t = 2 \, \text{s}$,

$$\vec{v} = \frac{2^2}{2} \, \hat{i} + 2^3 \, \hat{j} = 2 \, \hat{i} + 8 \, \hat{j}$$

Also, at $t = 2 \, \text{s}$,

$$\vec{F} = 2 \, \hat{i} + 3 \cdot 2^2 \, \hat{j} = 2 \, \hat{i} + 12 \, \hat{j}$$

Power is

$$P = \vec{F} \cdot \vec{v}$$

So,

$$P = (2 \, \hat{i} + 12 \, \hat{j}) \cdot (2 \, \hat{i} + 8 \, \hat{j}) = (2 \cdot 2) + (12 \cdot 8) = 4 + 96 = 100 \, \text{W}$$

Therefore, the power at $t = 2 \, \text{s}$ is $100 \, \text{W}$.

Given: instantaneous power is found using the dot product of force and velocity.

Find: the numerical value of power at $t = 2 \, \text{s}$.

The alternate working states

$$P = \vec{F} \cdot \vec{V} = \frac{2t^3}{ } + 3t^5$$

Evaluating at $t = 2$ gives the final numerical result

$$P = 100$$

the solution explicitly concludes that the correct answer is $100$.

There is a discrepancy between the question text and the force expression used in the solution. The answer has been derived from the solution, which is the primary source here.