Question

Sand from a stationary hopper falls onto a moving conveyor belt at the rate of 5.00 kg/s. The conveyor belt is supported by frictionless rollers and moves at a constant speed of v = 0.75 m/s under the action of a constant horizontal external force Fext supplied by the motor that drives the belt. Find (a) the sand’s rate of change of momentum in the horizontal direction, (b) the external force Fext (c) the work done by Fext in 1 s, and (e) the kinetic energy acquired by the falling sand each second due to the change in its horizontal motion.

Answer 1

$a)\frac{m(V-V_0)}{t}=\frac{mV}{t}=\frac{5\times 0.75}{1}=3 \frac{kg\times m}{s^2}=3N;\\b)\frac{m(V-V_0)}{t}=F_{ext}=3N;\\c)\frac{A}{t}=\frac{F_{ext}S}{t}=F_{ext}V =3\times0.75=2.25\frac{J}{ s}\\or\\\ P=2.25W ; \\e)A=\Delta E_k;\Delta E_k=2.25J$