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 as in Figure P9.72.The conveyor belt is supported by frictionless rollers and moves ata constant speed of v = 0.740 m/s under the action of aconstant horizontal external force Fextsupplied by 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 forceFext. (d) Find the work done by Fext in1 second.
(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 as in Figure P9.72.The conveyor belt is supported by frictionless rollers and moves ata constant speed of v = 0.740 m/s under the action of aconstant horizontal external force Fextsupplied by 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 forceFext. (d) Find the work done by Fext in1 second.
(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

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.