Sand from a stationary hopper falls on a moving conveyor belt at the rate of 4.2

Question

Sand from a stationary hopper falls on a moving conveyor belt at the rate of 4.20 kg/s as in the figure below. The conveyor belt is supported by frictionless rollers and moves at a constant speed of v 0.620 m/s under the action of a constant horizontal external force Fext supplied by the motor that drives the belt. (a) Find the sand's rate of change of momentum in the horizontal direction. (b) Find the force of friction exerted by the belt on the sand. (c) Find the external force Fext (d) Find the work done by Fext in 1 second. (e) Find the kinetic energy acquired by the falling sand each second due to the change in its horizontal motion

Explanation / Answer

ANSWER :-

a) The mass of sand falling on the belt in time dt is

1) dm = dm/dt * dt

where dm/dt is the given rate at which the sand fall of the conveyor belt.

we can assume that the initial momentum ofsand fallen on the belt in time dt has a zero value.

2) dPx/dt = 0 dm = v x dm/ dt *dt

where vx is v x dm/dtthe speed of the conveyor belt. Hence the rate of change inthe total momentum of the sand is

3) dPx/dt = v x dm/dt = 4.20 * 0.630 = 2.646N

b) There are three forces exerted on the sand but only the frictionalforce is not balanced. The net force is therefore equal to the frictionalforce. Hence, from Newton’s second law of motion, the x-component of the (total) frictional force exerted on the sand is 2.65 N.

c) From Newton’s third law of motion, the sand exerts an opposite,frictional force with the x-component of -2.65 N. According to Newton’ssecond law of motion, the net force exerted on the conveyor belt is zero,which requires that the motor exert a force with the x-component of 2.65 N.

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