JEE Physics 2023 Question with Solution

Given: $\mathbf{F} = (2 + 3x)\hat{i}$, displacement from $x=0$ to $x=4 \, \text{m}$.

Find: Work done by the variable force.

For a variable force along the $x$-direction, work done is

$$W = \int_{x_1}^{x_2} F(x) \, dx$$

Here,

$$W = \int_0^4 (2+3x) \, dx$$

Now split the integral:

$$W = \int_0^4 2 \, dx + \int_0^4 3x \, dx$$ $$W = [2x]_0^4 + \left[\frac{3x^2}{2}\right]_0^4$$

Substituting the limits:

$$W = (2 \times 4) + \left(\frac{3 \times 4^2}{2}\right)$$ $$W = 8 + \left(\frac{3 \times 16}{2}\right) = 8 + 24 = 32 \, \text{J}$$

Therefore, the work done is $32 \, \text{J}$. The correct option is A.

Given: $F(x)=2+3x$.

Find: Work done from $x=0$ to $x=4$.

Take the antiderivative directly:

$$W = \int_0^4 (2+3x) \, dx = \left[2x + \frac{3x^2}{2}\right]_0^4$$

Now evaluate at the limits:

$$W = \left(2\cdot 4 + \frac{3\cdot 16}{2}\right) - 0 = 8 + 24 = 32 \, \text{J}$$

This works because work by a variable force is the area under the $F$-versus-$x$ curve. Therefore, the correct option is A.