Question

1. A 1000 kg equipment lift travels up a 200 m shaft in 45 seconds. Assuming the speed is constant, what is the
average power?

Answer 1

Answer:

43,555 W

Explanation:

Since the speed is constant, this means that the force applied to lift the equipment is equal to the weight of the equipment. So we can write:

$F=mg$

where

m = 1000 kg is the mass

$g=9.8 m/s^2$ is the acceleration due to gravity

So,

$F=(1000)(9.8)=9800 N$

Then the work done in lifting the equipment is:

$W=Fd$

where

d = 200 m is the displacement of the equpment

Substituting,

$W=(9800)(200)=1.96\cdot 10^6 J$

Finally, the power used to lift the equipment is the ratio between the work done and time taken:

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

where

$W=1.96\cdot 10^6 J$

t = 45 s is the time taken

Solving,

$P=\frac{1.96\cdot 10^6}{45}=43,555 W$