Sand from a stationary hopper falls on amoving conveyor belt at the rate of 5.70

Question

Sand from a stationary hopper falls on amoving conveyor belt at the rate of 5.70 kg/s. The conveyor belt issupported by frictionless rollers and moves at a constant speed ofv = 0.740 m/s under the action of a constanthorizontal external force Fext suppliedby the motor that drives the belt. (a) Find the sand's rate of change of momentum in thehorizontal direction. (b) Find the force of friction exerted by the belt on thesand.
(c) Find the external force Fext.
(d) Find the work done by Fext in 1second.
(e) Find the kinetic energy acquired by the falling sand eachsecond due to the change in its horizontal motion.
(f) Why are the answers to parts (d) and (e) different?
Sand from a stationary hopper falls on amoving conveyor belt at the rate of 5.70 kg/s. The conveyor belt issupported by frictionless rollers and moves at a constant speed ofv = 0.740 m/s under the action of a constanthorizontal external force Fext suppliedby the motor that drives the belt. (a) Find the sand's rate of change of momentum in thehorizontal direction. (b) Find the force of friction exerted by the belt on thesand.
(c) Find the external force Fext.
(d) Find the work done by Fext in 1second.
(e) Find the kinetic energy acquired by the falling sand eachsecond due to the change in its horizontal motion.
(f) Why are the answers to parts (d) and (e) different?

Explanation / Answer

   from the given problem we can see that the sand isfalling continuosly and the stationary hopper
   is moving with a velocity v = 0.740 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) (0.740m / 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 (0.740 m)cos0o
        = ......... J
(e)
   the kinetic energy is given by
   (1 / 2) (m) v2 = (1 / 2) (4.61kg) (0.740 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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