Afully loaded, slow-moving freight elevator has a cab with a total mass of 1250 kg, which is required to travel upward 53 m in 3.1 min, starting and ending at rest. the elevator's counterweight has a mass of only 975 kg, and so the elevator motor must . what average power is required of the force the motor exerts on the cab via the cable?

P= 783.6 W

Explanation:

Given that

Total mass M= 1250 Kg

d= 53 m ,in time t= 3.1 min

Counter weight mass ,m = 975 kg

The total net force F

F=  M g - m g

F= (1250 - 975 ) x 10   N             ( take g =10 m/s²)

F=2750 N

We know that work done by force F

W= F. d

W= 2750 x 53 J

W=145750 J

The power P is the rate of change of work with time .

P= 783.6 W

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s