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3 years ago

A 500 N passenger weighs 24500 N elevator that rises 30 m in exactly 1 minute how much power is needed for the elevators trip

Physics

1 answer:

larisa [96]3 years ago

5 0

Answer:

12500 W (12.5 kW)

Explanation:

The total weight of the passenger + elevator is

$F_g = 500 N + 24500 N=25000 N$

The total work done to rise the elevator + passenger is equal to the product between the total weight and the distance covered during the trip (d = 30 m):

$W=F_g d = (25000 N)(30 m)=750,000 J$

The power needed for the trip is equal to the ratio between the work done (W) and the time taken (t):

$P=\frac{W}{t}$

Since the time taken is t = 1 min = 60 s, the power needed is

$P=\frac{750,000 J}{60 s}=12,500 W=12.5 kW$

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