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?


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.




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