Sand from a stationary hopper falls on a moving conveyor belt at the rate of 5.3

Question

Sand from a stationary hopper falls on a moving conveyor belt at the rate of 5.30 kg's as in the figure below. The conveyor belt is supported by frictionless rollers and moves at a constant speed of v 0.630 m/s under the action of a constant honzontal 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. (e) rind the external force Fext (d) Find the work done by Fext in 1 second. ) Find the kinctic energy acquired by the falling sand cach sccond duc to the change in its horizontal motion (f) Why are the answers to parts (d) and (e) different?

Explanation / Answer

Given, velocity v = 0.63 m /s

mass, m = 5.30 kg of sand moving on the belt

(a) the sand's rate of change of momentum in thehorizontal direction will be

   ?px / ?t = (5.3 kg) (0.63 m / s) / (1.00 s) = 3.34 N

(b) from the given we can see that the onlyhorizontal force on the sand is the belt friction

   so that, pxi + f ?t =pxf

   f = ?px / ?t = 3.34 N

(c) when the belt is in equilibrium

   ? Fx = m ax

   Fext - f = 0

   Fext = f = 3.34 N

(d) the work done will be

   W = F ?r cos?

        = F (0.63 m)cos0o = 0.63 J

(e) the kinetic energy is given by

   (1 / 2) (?m) v2 = (1 / 2)(5.3 kg) (0.63 m / s)2

                           = 1.05 J

(f) the firction between the sand and the belt converts half of the input work into extra internal

   energy

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