Work Done Against Friction

Because the sled moves at constant velocity, its kinetic energy does not change, so by the work-energy theorem the net work is zero and the pulling force must exactly balance kinetic friction. The frictional force is $f=\mu mg=0.2\times 10\times 10=20\text{ N}$, so the horizontal pull is also 20 N. The work done by this pull over 25 m is $W=Fd=20\times 25=500\text{ J}$. The option 250 J halves the distance contribution without cause. The option 2500 J inflates the friction force by a factor of five through a coefficient error. The option 50 J drops the displacement to a tenth. The crucial conceptual point is that constant velocity means zero acceleration and therefore zero net force, so the rope tension and the friction force are equal and opposite at every instant of the drag. Unlike work done against gravity, this energy cannot be recovered later, because it degrades irreversibly into thermal energy spread across the contact surface. This reflects the NCERT idea that at constant speed applied work equals the energy dissipated by friction. As a sanity check, all 500 J becomes heat at the contact, since the kinetic energy of the sled is unchanged from start to finish.