JEE Physics 2025 Question with Solution

Given: Mass of block is $m = 1 \, \text{kg}$, initial speed is $v_i = 10 \, \text{m/s}$, rough region extends from $x_i = 0.1 \, \text{m}$ to $x_f = 1.9 \, \text{m}$, and retarding force is $F_r = -kx$ with $k = 10 \, \text{N/m}$.

Find: Final speed $v_f$ of the block after crossing the rough region.

Use the work-energy theorem. The work done by the retarding force is

$$W = \int_{x_i}^{x_f} F_r \, dx = \int_{0.1}^{1.9} (-kx) \, dx$$

Substituting $k = 10 \, \text{N/m}$,

$$W = \int_{0.1}^{1.9} (-10x) \, dx = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9}$$ $$W = -5 \left[ (1.9)^2 - (0.1)^2 \right] = -5 [3.61 - 0.01] = -5 [3.60] = -18 \, \text{J}$$

Now apply the work-energy theorem:

$$W_{\text{net}} = \Delta KE = KE_f - KE_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$$

Since the retarding force does the net work in this region,

$$-18 = \frac{1}{2} (1) v_f^2 - \frac{1}{2} (1) (10)^2$$ $$-18 = \frac{1}{2} v_f^2 - 50$$ $$\frac{1}{2} v_f^2 = 32$$ $$v_f^2 = 64$$ $$v_f = \sqrt{64} = 8 \, \text{m/s}$$

Therefore, the final speed of the block as it crosses the rough region is $8 \, \text{m/s}$. The correct option is D.

Given: $m = 1 \, \text{kg}$, $v_i = 10 \, \text{m/s}$, $F_r = -kx$, $k = 10 \, \text{N/m}$, and motion through the interval $0.1 \, \text{m} \le x \le 1.9 \, \text{m}$.

Find: The final speed $v_f$.

The work done by the retarding force is

$$W = \int_{0.1}^{1.9} -kx \, dx$$ $$W = -k \left[ \frac{x^2}{2} \right]_{0.1}^{1.9}$$ $$W = -10 \left[ \frac{1.9^2}{2} - \frac{0.1^2}{2} \right]$$ $$W = -10 \left[ \frac{3.61}{2} - \frac{0.01}{2} \right]$$ $$W = -10 [1.805 - 0.005]$$ $$W = -10 \times 1.8 = -18 \, \text{J}$$

Now,

$$\Delta KE = KE_f - KE_i = W$$ $$\frac{1}{2} m v_f^2 - \frac{1}{2} \times 1 \times 10^2 = -18$$ $$\frac{1}{2} \times 1 \times v_f^2 - 50 = -18$$ $$\frac{1}{2} v_f^2 = 32$$ $$v_f^2 = 64$$ $$v_f = \sqrt{64} = 8 \, \text{m/s}$$

Therefore, the final speed is $8 \, \text{m/s}$.