Question #99564

Sand drops vertically at the rate of 2kg/s unto conveyor belt moving horizontally with a velocity of 0.1m/s.
a) Calculate the extra power needed to keep the belt moving.
b) The rate of change of kinetic energy of the sound.
,c) While is the power twice as rate as the rate of change of kinetic energy.

Expert's answer

a)


F=vdmdtF=v\frac{dm}{dt}

P=Fv=v2dmdt=(0.1)22=0.02 WP=Fv=v^2\frac{dm}{dt}=(0.1)^22=0.02\ W

b)

dKdt=0.5v2dmdt=0.5(0.1)22=0.01 W\frac{dK}{dt}=0.5v^2\frac{dm}{dt}=0.5(0.1)^22=0.01\ W

c) The power is the rate of change of work, not kinetic energy.



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