At a certain factory, 270 kg crates are dropped vertically from a packing machin

Question

At a certain factory, 270 kg crates are dropped vertically from a packing machine onto a conveyor belt moving at 1.20 m/s. (A motor maintains the belt's constant speed.) The coefficient of kinetic friction between belt and crate is 0.400. After a short time, slipping between the belt and the crate ceases and the crate then moves along with the belt. Consider the period of time during which the crate is being brought to rest relative to the belt.

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(a) Calculate, for a coordinate system at rest in the factory, the kinetic energy supplied to the crate.


(b) Calculate, for a coordinate system at rest in the factory, the magnitude of the kinetic frictional force acting on the crate.


(c) Calculate, for a coordinate system at rest in the factory, the energy supplied by the motor.



(d) Explain why the answers to (a) and (c) are different.

Explanation / Answer

      The mass of the crate, m = 270 kg       The speed of the conveyor belt, v = 1.20 m/s       The coefficient of kinetic friction, k = 0.400 _______________________________________________________________ a)       The kinetic energy supplied to the crate is                     K.E = (1/2)mv2                            = (1/2)(270 kg)(1.20 m/s)2                            = 194.4 J _______________________________________________________________ _______________________________________________________________ b)       The magnitude of the kinetic frictional force acting on the crate is                           fk= kN                                   = kmg                                = (0.400)(270 kg)(9.8 m/s2)                                = 1058.4 N ________________________________________________________________ ________________________________________________________________ c)       The energy supplied by the motor is                         W = E+(K.E)                             = fkd +(K.E)
                            = (1/2)mv2+(K.E)                             = (K.E) +(K.E)                             = 2(K.E)                             = 2(194.4 J)
                            = 388.8 J       ________________________________________________________________ ________________________________________________________________ d)       The ansewer of part (a) and part (c) are different,       because the energy supplied by the motor (in part c) is       the sum of the thermal energy and kinetic energy supplied to the crate.                                     = (0.400)(270 kg)(9.8 m/s2)                                = 1058.4 N ________________________________________________________________ ________________________________________________________________ c)       The energy supplied by the motor is                         W = E+(K.E)                             = fkd +(K.E)
                            = (1/2)mv2+(K.E)                             = (K.E) +(K.E)                             = 2(K.E)                             = 2(194.4 J)
                            = 388.8 J       ________________________________________________________________ ________________________________________________________________ d)       The ansewer of part (a) and part (c) are different,       because the energy supplied by the motor (in part c) is       the sum of the thermal energy and kinetic energy supplied to the crate.     

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