A)A 675-kg elevator starts from rest andmoves upward for 3.40 s with constantacc

Question

A)A 675-kg elevator starts from rest andmoves upward for 3.40 s with constantacceleration until it reaches its cruising speed, 1.65 m/s. (1) What is the average power of the elevatormotor during this period?
1 W

(2) How does this amount of power compare with its power during anupright trip with constant speed?
2 W


B) A 21.0 kg child on a 3.00 m long swing is released from rest when the ropesof the swing make an angle of 26.0° with the vertical. (1) Neglecting friction, find the child's speedat the lowest position.
1 m/s

(2) If the speed of the child at the lowest position is2.10 m/s, what is the mechanical energylost due to friction?
2 J
(1) What is the average power of the elevatormotor during this period?
1 W

(2) How does this amount of power compare with its power during anupright trip with constant speed?
2 W


B) A 21.0 kg child on a 3.00 m long swing is released from rest when the ropesof the swing make an angle of 26.0° with the vertical. (1) Neglecting friction, find the child's speedat the lowest position.
1 m/s

(2) If the speed of the child at the lowest position is2.10 m/s, what is the mechanical energylost due to friction?
2 J
(1) Neglecting friction, find the child's speedat the lowest position.
1 m/s

(2) If the speed of the child at the lowest position is2.10 m/s, what is the mechanical energylost due to friction?
2 J

Explanation / Answer

Force of power motor - mg = ma .         F  - mg = m v / t .         F = mv/t + mg = 675*1.65/3.40  + 675*9.8   =   6942.57 . Power = force * avg velocity = 6942.57* (1/2) * 1.65   =    5727 Watts . Power at constant speed = mgv = 675 * 9.8 *1.65 =  10915 Watts

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