1. Given data

   • Mass of block: $m = 100\,\text{kg}$
   • Distance moved: $s = 10\,\text{m}$
   • Coefficient of friction: $\mu = 0.4$
   • Surface is horizontal, so normal reaction is $N = mg$

2. Frictional force The friction force is $f = \mu N = \mu mg$ Substituting values: $f = 0.4 \times 100 \times 9.8 = 392\,\text{N}$

3. Work done against friction Work done against friction over distance $s$ is $W = f s$ $W = 392 \times 10 = 3920\,\text{J}$

4. Choosing the nearest option The exact value using $g=9.8\,\text{m/s}^2$ is $3920\,\text{J}$ In JEE-type problems, we usually take $g = 10\,\text{m/s}^2$ unless otherwise specified. Then, $f = 0.4 \times 100 \times 10 = 400\,\text{N}$ $W = 400 \times 10 = 4000\,\text{J}$

5. Option check

   • A: $3900$ ❌
   • B: $4500$ ❌
   • C: $4200$ ❌
   • D: $4000$ ✅

Therefore, the correct answer is Option D.