Answer to Question #141196 in Mechanics | Relativity for Andy

Question

Sand is deposited at a uniform rate of 20 Kg/s and with negligible kinetic energy on to an empty conveyor belt moving horizontally at a constant speed of 10 meters per minute. Find (a) the force required to maintain the constant velocity, (b) the power required to maintain a constant velocity, and (c) the rate of change of kinetic energy of the moving sand

Answer 1

From Newton's second law, Force is defined as the rate of change in momentum.

$F=\frac{dP}{dt}=\frac{d(mv)}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}\\ F=m\frac{dv}{dt}+v\frac{dm}{dt}\hspace{2cm}(1)\\ \textsf{where P, m, v and t represent momentum,mass, velocity and time respectively.}\\$

(a) Since the conveyor belt is to maintain a constant velocity, $\frac{dv}{dt}=0.$

Equation (1) therefore becomes $F=v\frac{dm}{dt}.\\ \textsf{From the question,}\\ v=\frac{10m}{min} = \frac{10m}{60s}= \frac{1}6ms^{-1}\\ \frac{dm}{dt}=20kgs^{-1}\\\hspace{2cm}\\ \textsf{The Force is,}\\ F=\frac{1}6ms^{-1} × 20kgs^{-1}=\frac{10}3kgms^{-2}\\ F=3.33N.$

(b) The power is given by,

$P=F×v P=3.33N×\frac{1}6ms^{-1}\\ P=0.56W$

(c) From work energy principle,

Change in Kinetic energy = Workdone

$\frac{d(K.E)}{dt}=\frac{Workdone}{time}\\ \frac{d(K.E)}{dt}=Power\\ \frac{d(K.E)}{dt}=0.56W$

The rate of change of kinetic energy of the moving sand is therefore $0.56W$