Sand is being dropped from a stationary dropper at a rate of 0.5 kgs⁻¹ on a conveyor belt moving with a velocity of 5 ms⁻¹. The power needed to keep the belt moving with the same velocity will be :

Solution

Physics sir · 4 steps

The belt has to push every new kilogram of sand up to 5 m/s, so find that force first, then multiply by v.

1. Force on the belt. Sand lands with zero horizontal speed and leaves moving at v. The belt must supply momentum at the rate v dm/dt, so F = 5 × 0.5 = 2.5 N.

   P = Fv = v² dm/dt

2. Put in numbers. P = (5)² × 0.5 = 25 × 0.5 = 12.5 W.
3. Where it goes. Only half of this, 1/2v²dm/dt = 6.25 W, becomes kinetic energy of the sand. The other half is lost as heat while the sand slides on the belt before it catches up to the belt's speed.
4. So. The power needed is 12.5 W, option (d).

The idea to keep. For mass landing on a moving belt, use F = v dm/dt and P = Fv. Do not use the kinetic energy gain, because friction wastes an equal amount as heat.

Answer from the NTA final answer key. Solution by PrepWiser.