JEE Main 2025 — Laws Of Motion Question with Solution

Work Done = Final Kinetic Energy - Initial Kinetic Energy

$$W=\Delta K=K_f-K_i$$

Calculate the Work Done by the Retarding Force

The problem gives a retarding force $$F_r=-kr$$. It seems 'r' is a typo for 'x', the position, since the constant $$k$$ has units of N/m. So, the force is a function of position: $$F_r=-kx$$.

Because the force isn't constant, we need to find the work done by integrating the force over the distance the block travels in the rough region (from $$x=0.1\text{ m}$$ to $$x=1.9\text{ m}$$).

$$W={\int }_{0.1}^{1.9}{F}_{r}dx={\int }_{0.1}^{1.9}(-kx)dx$$

Plugging in $$k=10\text{ N/m}$$:

$$W=-10{\int }_{0.1}^{1.9}xdx$$

$$W=-10{\left[\frac{x^2}{2}\right]}_{0.1}^{1.9}$$

$$W=-5\left[(1.9{)}^2-(0.1{)}^2\right]$$

$$W=-5[3.61-0.01]=-5(3.6)$$

$$W=-18\text{ J}$$

The negative sign just means the force did negative work, slowing the block down.

Calculate the Initial Kinetic Energy ($$K_i$$)

The initial kinetic energy is found using the formula $$K_i=\frac{1}{2}m{v}_i^2$$.

Given:

Mass $$m=1\text{ kg}$$

Initial speed $$v_i=10\text{ m/s}$$

$$K_i=\frac{1}{2}(1\text{ kg})(10\text{ m/s}{)}^2=\frac{1}{2}(100)=50\text{ J}$$

Use the Work-Energy Theorem to find the Final Speed ($$v_f$$)

Now we can put everything together into the Work-Energy Theorem equation: $$W=K_f-K_i$$.

We know $$W=-18\text{ J}$$ and $$K_i=50\text{ J}$$. The final kinetic energy is $$K_f=\frac{1}{2}m{v}_f^2$$.

$$-18=\left(\frac{1}{2}(1)v_f^2\right)-50$$

Now, let's solve for $$v_f$$:

$$-18+50=\frac{1}{2}{v}_f^2$$

$$32=\frac{1}{2}{v}_f^2$$

$$v_f^2=64$$

$$v_f=\sqrt{64}$$

$$v_f=8\text{ m/s}$$

The final speed of the block as it leaves the rough region is 8 m/s.

Therefore, the correct option is C.