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 conveyor beltis supported by frictionless rollers and moves at a constant speedof v = 0.740 m/s under the action of a constant horizontalexternal force Fext supplied by themotor 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 forceFext.
(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.

Explanation / Answer

   from the given problem we can see that the sandis falling continuosly and the stationary hopper
   is moving with a velocity v = 0.740 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)(0.740 m / s) / (1.00 s)
               = ........ N  
(b)
from the given we can see that the only horizontal force onthe 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)(5.70 kg) (0.740 m / s)2                            =......... J (f)    the firction between the sand and the beltconverts half of the input work into extra internal    energy

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