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

43,555 W

Step-by-step 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=(W)/(t)`

where

`W=1.96\cdot 10^6 J`

t = 45 s is the time taken

Solving,

`P=(1.96\cdot 10^6)/(45)=43,555 W`