Question

Sand from a stationary hopper falls on a moving conveyor belt at the rate of 4.60 kg/s as in fig. P9.72. The conveyor belt is supported be frictionless rollers and moves at a constant speed of v = 0.820 m/s under the action of a consant horizontal external force Fext supplied by the motor that drives the belt.

(a) Find the sand's rate of change of momentum in the hortizontal direction. N

(B) Find the force of friction exerted by the belt on the sand. N

(c) find the external force Fext. N

(d) Find the work done by Fext in 1 second. J

(e) Find the kinetic energy acquired by the falling sand each second due to the change in its hortizontal motion. J

(f) Why are the answers to parts (d) and (e) different?

Answer 1

(a) The sand's rate of change of momentum in the hortizontal direction which is given as :

using an equation,   dP_x / dt = (dm / dt) v                                                       { eq.1 }

where, dm/dt = rate at which the sand fall of the conveyor belt = 4.6 kg/s

v = speed of the conveyor belt = 0.82 m/s

inserting the values in above eq.

dP_x / dt = (4.6 kg/s) (0.82 m/s)

dP_x / dt = 3.77 N

(b) The force of friction exerted by the belt on the sand which is given as :

using an equation,    F_f = v (dm / dt)

F_f = (0.82 m/s) (4.6 kg/s)

F_f = 3.77 N

(c) The external force wil be given as :

using an equation, F_ext = v (dm / dt)

F_ext = (0.82 m/s) (4.6 kg/s)

F_ext = 3.77 N

(d) The work done by F_ext in 1 second which is given as :

using an equation,   W = F_ext v t                                                             { eq.2 }

inserting the values in eq.2,

W = (3.77 N) (0.82 m/s) (1 sec)

W = 3.09 J

(e) The kinetic energy acquired by the falling sand each second due to the change in its hortizontal motion which is given as :

K.E = (1/2) (dm / dt) v² t                                                             { eq.3 }

inserting the values in eq.3,

K.E = (0.5) (4.6 kg/s) (0.82 m/s)² (1 sec)

K.E = 1.54 J

(f) The answers to parts (d) and (e) are different because Change in the total kinetic energy of a system is related to work done by both external and internal interaction.

According to work-energy theorem, we have

W_internal = K.E_total - W_external                                                             { eq.4 }

inserting the values in eq.4,

W_internal = (1.54 J) - (3.09 J)

W_internal = -1.55 J