JEE Physics 2023 Question with Solution

Given: The block is pulled from rest to $x_0 = 10 \, \text{cm} = 0.1 \, \text{m}$ on a frictionless surface. At $x = 5 \, \text{cm} = 0.05 \, \text{m}$, the energy of the block is $0.25 \, \text{J}$.

Find: The spring constant $k$.

Initially, all the energy is spring potential energy:

$$U_i = \frac{1}{2} k x_0^2$$

The initial kinetic energy is zero because the block starts from rest.

At $x = 5 \, \text{cm} = x_0/2$, the spring potential energy is:

$$U_f = \frac{1}{2} k \left(\frac{x_0}{2}\right)^2 = \frac{1}{8} k x_0^2$$

The kinetic energy there is:

$$K_f = 0.25 \, \text{J}$$

Using conservation of mechanical energy:

$$U_i + K_i = U_f + K_f$$ $$\frac{1}{2} k x_0^2 + 0 = \frac{1}{8} k x_0^2 + 0.25$$ $$\frac{1}{2} k x_0^2 - \frac{1}{8} k x_0^2 = 0.25$$ $$\frac{3}{8} k x_0^2 = 0.25$$

So,

$$k = \frac{0.25 \times 8}{3 x_0^2} = \frac{2}{3 x_0^2}$$

Substitute $x_0 = 0.1 \, \text{m}$:

$$k = \frac{2}{3 (0.1)^2} = \frac{2}{3 \times 0.01} = \frac{2}{0.03} = \frac{200}{3} \approx 66.67 \, \text{N/m}$$

Rounding to the nearest integer gives $67 \, \text{N/m}$.

Therefore, the spring constant is $67 \, \text{N/m}$.

Given: Maximum displacement is $x_0 = 0.1 \, \text{m}$ and at $x = x_0/2$, the kinetic energy is $0.25 \, \text{J}$.

Find: The value of $k$.

For a spring-block system, total energy remains constant on a frictionless surface. At the extreme position, the entire energy is:

$$E = \frac{1}{2} k x_0^2$$

At half the displacement, the potential energy becomes:

$$U = \frac{1}{2} k \left(\frac{x_0}{2}\right)^2 = \frac{1}{8} k x_0^2$$

Hence the kinetic energy at that point is:

$$K = E - U = \frac{1}{2} k x_0^2 - \frac{1}{8} k x_0^2 = \frac{3}{8} k x_0^2$$

Given that:

$$\frac{3}{8} k x_0^2 = 0.25$$

Substituting $x_0 = 0.1 \, \text{m}$:

$$\frac{3}{8} k (0.1)^2 = 0.25$$ $$\frac{3}{8} k (0.01) = 0.25$$ $$0.00375k = 0.25$$ $$k = \frac{0.25}{0.00375} \approx 66.67$$

Thus, the required numerical value is $67$.