Question

The express elevator in the Sears Tower averages a speed of `10` `ms^-1` in its climb to the 103 floor, 410 meters above the ground. Assuming a total load of `1.0 * 10^3` kg, what average power must the lifting motor supply?

Answer 1

`P=mgv = 1.0*10^3*9.8*10=98000` `W`

See the explanation below for why this formula is appropriate.

Explanation:

Power is the rate at which work is done:

`P=W/t`

Mechanical work is the product of force and distance:

`W=Fd`

Substituting into the power expression:

`P=(Fd)/t`

But `d/t` is just another expression for velocity, so

`P=Fv`

But the force acting is `F=mg`, so:

`P=mgv = 1.0*10^3*9.8*10=98000` `W`

Note that we didn't actually need the height of the building, since this is about instantaneous power. We could have calculated this result a different way, using the total change in gravitational potential and the time taken for the lift to travel to the top.

Actually, why not do that as a check?

Potential energy:

`E_p=mgh=1*10^3*9.8*410=401800` `J`

This is the total work done, `W` `J`.

The time taken for the `410` `m` trip at `10` `ms^-1` is `41` `s`.

`P=W/t=4018000/41=98000` `W`

Happymaking when two different approaches yield the same answer. ;-)