Question

Sand drops vertically from a stationary hopper at a constant rate of 100 gram per second onto a horizontal conveyor belt moving at a constant velocity, ~v, of 10 cm/sec. 1. What force (magnitude and direction relative to the velocity) is required to keep the belt moving at a constant speed of 10 cm/sec? 2. How much work is done by this force in 1.0 second? 3. What is the change in kinetic energy of the conveyor belt in 1.0 second due to the additional sand on it? 4. Should the answers to parts 2. and 3. be the same? Explain.

Answer 1

Step-by-step explanation:

1 ) Force F = dp/ dt , p is momentum

= d/dt (mv)

= v dm/dt

Given dm/dt = 0.1 kg per second.

Force = v x 0.1

= .1 x .1

= .01 N

2 ) work done = force x displacement

Displacement in 1 s = 0. 1 m ( v = .1 m /s )

Work done in one second = .1 X .1

= .01 J

3 ) Velocity is constant so change in kinetic energy is due to additional mass .

Mass deposited in one second

= .1 g

Velocity = .1 m/s

Change in kinetic energy = 1/2 x .1 x .1 ²

= 5 x 10⁻⁴J

4 ) No , they are not the same. It is so because force has to do work against the frictional force offered by the surface on which sand is falling . So rise in kinetic energy is small as compared with work done by the force.