# Elevators and Potential

Hello everybody, i need some help!

i am trying to work out the energy required to move a lift from one floor to another with somebody in it who weighs 50kg.

The lift is going to be one mass, and it has a counterweight to help it go up and down.

thing to know:

the lift weighs 500kg on its own, when empty.

the distance between a floor is 2 metres

So if i need to know the energy to get it from floor 2 to 3, how would i calculate it? do i use potential energy, ie. mgh?

and what about when the lift is going downwards from floor 3 to floor 2, i wouldnt be able to use mgh because then it would be like it was falling, rather than being controlled

any help would be appreciated.

Mark44
Mentor
When the elevator is at floor 3, it has more potential energy than when it is at floor 2. The motor that runs the elevator has to do work to lift the elevator and the person in it to the next floor. The work done is the change in potential energy.

When the elevator goes down, it doesn't just fall down the elevator shaft: its descent is controlled. The motor and other parts of the system (brakes?) have to do work to keep the elevator moving at a constant speed.

tiny-tim
Homework Helper
Welcome to PF!

Hello MH03L! Welcome to PF! So if i need to know the energy to get it from floor 2 to 3, how would i calculate it? do i use potential energy, ie. mgh?

yes, use mgh (and don't forget to subtract mgh for the conuterweight )
and what about when the lift is going downwards from floor 3 to floor 2, i wouldnt be able to use mgh because then it would be like it was falling, rather than being controlled

remind me never to travel in a lift with you! energy is energy, you can always use mgh so from floor 2 to floor 3, the energy used by the lift will be the change in potential, i.e the work done?

so at floor 2:
mgh = 550 x 9.81 x 2 = 10791 J

at floor 3:
mgh = 550 x 9.81 x 4 = 21582 J

so the work done is:
21582 J - 10791 J = 10791 J

where does the counterweights and tension forces come into this?
thanks again!

thank you for the second reply too! much appreciated :)

and ahh so in my previous calculation i have

mgh = 550 x 9.81 x 2 = 10791 J for floor 2
but the counter weight weights 800kg.

so i have to take off
800 x 9.81 x h ??

my height will obviously be different though wont it. do i just guess where it is? haha
thanks again

tiny-tim
Homework Helper
Hi MH03L! where does the counterweights and tension forces come into this?

how heavy is the counterweight? (and tension is irrelevant)

how heavy is the counterweight?

hello tiny-tim :) thanks for responding
see my previous post, 800kg.

im still also confused about what happens whens its going downm because youre not going against gravity?

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Mark44
Mentor
the distance between a floor is 2 metres
That's a very unusual building.

lol! well they are very made up values, its just to get an understanding :)

tiny-tim
Homework Helper
my height will obviously be different though wont it.

no, it's attached to the lift over a pulley, it goes down the same distance the lift goes up im still also confused about what happens whens its going downm because youre not going against gravity?

it doesn't matter!!

if you go up you gain potential energy, if you go down you lose it

potential energy is defined as minus the work done …

go in the opposite direction, and the work done is negative Oh i see, well that makes sense. So if i take that off of my other calculation

mgh for the counterweight = 800 x 9.81 x 2 = 15696J
Then my mgh for the lift was 10791J, so the actual work done will be a negative for moving it upwards, that doesnt sound sensible??

thank you tiny-tim
Homework Helper
PE is minus the work done by gravity, that's plus the work done by you against gravity oh yes i see, sorry for being slow haha

to get the power for this then do you think i would just do >> work done/time

should power always be positive?? and if i want to know how much electrical power is used, is that just the number i calculate here, is it a straight conversion because its the same unit obviously

thanks tiny-tim
Homework Helper
to get the power for this then do you think i would just do >> work done/time

yes
should power always be positive??

no
and if i want to know how much electrical power is used, is that just the number i calculate here, is it a straight conversion because its the same unit obviously

no, it's a different unit

work and energy are in joules (J), power is in watts (W)

if you use SI units consistently (eg seconds rather than minutes), then there's no conversion factor

thanks :)

but my power = work done/ time can be used as a straight conversion ?

and also, going back to the travelling downwards (yes im still not getting it! haha)

to get work done it will be
800 x 9.81 x 2 ?? or will it be a negative acceleration ??

or, will it be the mgh at floor 3 minus the mgh at floor 2, which makes it negative?

tiny-tim
Homework Helper
1 watt = 1 joule per 1 second but can electrical power be negative?

Mark44
Mentor
but can electrical power be negative?
The question really should be "Can power be negative?"

Since Power = (Work done)/time, if the work done is negative, the power is negative.

cant you do the absolute value of it as surely same power would be needed to go from 1 to 2 as 2 to 1

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thankyou for the reply again :)

and ok so power can be negative, and i have a negative value for the lift when it goes up.

i have a positive power value for when the lift goes down.

but i want to know how much electrical energy is needed in total? obviously i dont add them because it will give a small value. do i maybe change the negative one to positive for electrical energy needed?

thanks again :)

and i thought it would be different from floor 1 to floor 2 than floor 2 to floor 1, is it the same?

Mark44
Mentor
The elevator consumes energy whether it goes up or down. If the motor does work in raising the elevator to a higher floor, the elevator does work on something when it comes down. We don't know any of the details of this elevator, but something -- brakes, the motor, whatever -- has to apply a force upward to keep the elevator from accelerating downward when it is moving to a lower floor.

okay thank you!

im still a bit confused on one thing though

i have worked out my PE mgh to be 29430 on the 3rd floor

and on the 1st floor i have worked it out to be 9810.

If the lift was going down would i say the work done is
29430-9810

and when the lift goes up the work done is
9810-29430

or should it be the other way around?
thanks for all your help i really appreciate it

Mark44
Mentor
okay thank you!

im still a bit confused on one thing though

i have worked out my PE mgh to be 29430 on the 3rd floor

and on the 1st floor i have worked it out to be 9810.

If the lift was going down would i say the work done is
29430-9810

and when the lift goes up the work done is
9810-29430
In one direction, the elevator motor will do work to move the elevator. In the other direction, the elevator will do work on the motor, brakes, etc. It's possible that with only one person in the elevator, the motor might have to do work to lift the counterweight (i.e., lower the elevator). There's not enough information given to know if the counterweight is equal to the weight of an unloaded elevator compartment or one loaded to capacity.
or should it be the other way around?
thanks for all your help i really appreciate it