JEE Physics 2025 Question with Solution

Given: Mass of the block is $m = 2 \, \text{kg}$, spring constant is $k = 200 \, \text{N/m}$, natural length of the spring is $2 \, \text{m}$, and the spring is initially compressed to length $1 \, \text{m}$.

Find: The speed of the block when it is at distance $x$ from the wall.

Working could not be extracted because the solution is unavailable.

Using conservation of mechanical energy, the initial compression from natural length is $1 \, \text{m}$, so the initial spring energy is

$$\frac{1}{2}k(1)^2 = \frac{1}{2}(200)(1)^2 = 100 \, \text{J}$$

When the block is at distance $x$ from the wall, the spring deformation is $2 - x$, so the total energy relation is

$$\frac{1}{2}mv^2 + \frac{1}{2}k(2-x)^2 = 100$$

Substituting $m = 2$ and $k = 200$,

$$v^2 + 100(2-x)^2 = 100$$ $$v^2 = 100\left[1-(2-x)^2\right]$$ $$v = 10\left[1-(2-x)^2\right]^{1/2}$$

Therefore, the correct option is C.