Sand from a stationary hopper falls on a moving conveyor beltat the rate of 5.70

Question

Sand from a stationary hopper falls on a moving conveyor beltat the rate of 5.70 kg/s. The conveyorbelt is supported by frictionless rollers and moves at a constantspeed of v = 0.690 m/s underthe action of a constant horizontal external force Fextsupplied by the motor that drives the belt. (a) Find the sand's rate of change of momentum in thehorizontal direction.
1 N

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

(c) Find the external force Fext.
3 N

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

(e) Find the kinetic energy acquired by the falling sand eachsecond due to the change in its horizontal motion.
5 J

(f) Why are the answers to parts (d) and (e) different? Sand from a stationary hopper falls on a moving conveyor beltat the rate of 5.70 kg/s. The conveyorbelt is supported by frictionless rollers and moves at a constantspeed of v = 0.690 m/s underthe action of a constant horizontal external force Fextsupplied by the motor that drives the belt. (a) Find the sand's rate of change of momentum in thehorizontal direction.
1 N

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

(c) Find the external force Fext.
3 N

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

(e) Find the kinetic energy acquired by the falling sand eachsecond due to the change in its horizontal motion.
5 J

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

Explanation / Answer

   from the given problem we can see that the sandis falling continuosly and the stationary hopper
   is moving with a velocity v = 0.690 m / s
   so after one second the stationary hopper differs byshowing 5.70 kg of sand moving on the belt
(a)
   the sand's rate of change of momentum in thehorizontal direction will be
   px / t = (5.70 kg)(0.690 m / s) / (1.00 s)
                = ........ N
(b)
   from the given we can see that the only horizontalforce on the sand is the belt friction
   so that
   pxi + f t = pxf
   f = px / t
     = ........ N
(c)
   when the belt is in equilibrium
   Fx = m ax
   Fext - f = 0
   Fext = f
          = ........N
(d)
   the work done will be
   W = F r cos
        = F (0.690 m)cos0o
        = ......... J
(e)
   the kinetic energy is given by
   (1 / 2) (m) v2 = (1 / 2) (5.70kg) (0.690 m / s)2
                          = ......... J
(f)
   the firction between the sand and the belt convertshalf of the input work into extra internal
   energy

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