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A fully 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 help. What average power is required of the force the motor exerts on the cab via the cable?

Respuesta :

Answer:

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 .

[tex]P=\dfrac{W}{t}[/tex]

[tex]P=\dfrac{145750}{3.1\times 60}\ W[/tex]

P= 783.6 W

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