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2 years ago

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A(n) 591 kg elevator starts from rest. It moves upward for 4.79 s with a constant acceleration until it reaches its cruising spe

ed of 2.17 m/s. The acceleration of gravity is 9.8 m/s 2 . Find the average power delivered by the elevator motor during the period of this acceleration. Answer in units of kW

1 answer:

Cloud [144]2 years ago

3 0

Answer:

Power delivered will be 10755.87 W

Explanation:

We have given mass of elevator m = 591 kg

Time t = 4.79 sec

As the elevator starts from rest so initial velocity u = 0m/sec

Final velocity v = 2.17 m/sec

From first equation of motion

v=u+at. here v is final velocity, u is initial velocity and t is time

So $2.17=0+a\times 4.79$

$a=0.453m/sec^2$

Height is given by $h=ut+\frac{1}{2}at^2=0\times 4.79+\frac{1}{2}\times 0.453\times 4.79^2=5.196m$

As elevator moves upward and acceleration due to gravity is downward so net acceleration due to gravity is downward so net acceleration = g-a=9.8-0.453 = 9.347 $m/sec^2$

Now work done $W=m(g-a)h=951\times (9.8-0.453)\times 5.196=51520.626J$

We know that power $P=\frac{work\ done}{time}=\frac{51520.626}{4.79}=10755.871W$

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