## Question:

Sand falls vertically at a rate of 50 kg ${{s}^{-1}}$ onto a horizontal conveyor belt moving at a speed of 1 m ${{s}^{-1}}$, as shown in the figure. What is the minimum power output of the engine which drives the belt? How is the work done by the engine accounted for?

## Step-by-step

A volume of sand of mass ∆m = 50 kg reaches a speed of $v=1m{{s}^{-1}}$ in time ∆t = 1 s. The change in its horizontal momentum is, therefore, $∆p = ∆mv = 50 kg m {{s}^{-1}}$. This means that a force $F={\frac {\Delta p} {\Delta t}}={\frac {v\Delta m} {\Delta t}}=50N$ accelerates the sand. The work done by the engine – taking only the acceleration of the sand into account – is $50 J {{s}^{-1}}$, i.e. its power output is 50 W. The sand loses its vertical momentum when it lands on the conveyor belt. (It hits the belt vertically with a force greater than its weight.) The sand on the belt is slowed down vertically and then accelerated horizontally; the belt’s kinetic energy is increased. The kinetic energy of the sand increases by $∆m {{v}^{2}}/2 = 25 J$ in ∆t = 1 s. This means that one-half of the power of the engine (25 W) is converted into kinetic energy of the sand; the rest is the work done against friction and converted into heat.
Note. The average speed of the sand during the acceleration is v/2. Therefore the power of the frictional force (F = 50 N) is Fv/2 = 25 W. The belt experiences a force −F, the power of which is −Fv = −50 W. Thus, exactly one-half of the power of the engine is used to accelerate the sand.