JEE Physics 2023 Question with Solution

Given: A block of mass $10\,kg$ moves under the force $F(x) = 5x$ from $x_1 = 2\,m$ to $x_2 = 4\,m$.

Find: The work done by the force.

For a position-dependent force, work is

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

Substituting the given force,

$$W = \int_{2}^{4} 5x\,dx$$ $$W = 5\int_{2}^{4} x\,dx = 5\left[\frac{x^2}{2}\right]_{2}^{4}$$

Applying the limits,

$$W = 5 \times \frac{1}{2}(4^2 - 2^2)$$ $$W = \frac{5}{2}(16 - 4) = \frac{5}{2} \times 12 = 30\ J$$

Therefore, the work done is $30\ J$.

Given: $F(x) = 5x$ and displacement from $x = 2$ to $x = 4$.

Find: Work done.

Since force depends linearly on position, integrate directly:

$$W = \int_{2}^{4} 5x\,dx = \frac{5}{2}\left[x^2\right]_{2}^{4}$$ $$W = \frac{5}{2}(16 - 4) = 30\ J$$

This works because the area under the $F$ versus $x$ graph gives the work done. Hence, the answer is $30\ J$.