What is the Average Power of an Elevator in Physics?

[MatinSAR]

Homework Statement

An electric elevator lifts a 20kg body 40m at an average speed of 2m/s.(The motion is upward.)
Average power of the elevator?

Relevant Equations

$P_{av}=Work/Time$

The answer in the book is 400w. It said that $P_{av}=Fv_{av}cos(F,v)=mgv_{av}cos(F,v)=400$w and F is upward force that is applied by elevator.
Should velocity of the elevator be constant? Because it said that we have F=mg.

 

[PeroK]

Should velocity of the elevator be constant? Because it said that we have F=mg.

That's the force required to move an object of mass $m$ upwards at constant speed. Isn't it?

 

[MatinSAR]

That's the force required to move an object of mass $m$ upwards at constant speed. Isn't it?

Yes. But in question it wasn't mentioned that speed is constant. It gives us average speed.
In asnwer it said that F=mg and because of this I thought that speed is constant.

 

[PeroK]

Yes. But in question it wasn't mentioned that speed is constant. It gives us average speed.
In asnwer it said that F=mg and because of this I thought that speed is constant.

Does the work done depend on the speed profile?

 

[MatinSAR]

Does the work done depend on the speed profile?

Yes. According to Work-Energy theorem Work depends on change in speed.

 

[MatinSAR]

If we assume that speed is constant the total work is 0 and again it shows that F=mg.

The problem is that if we assume that speed is not constant then F≠mg ...Update:

According to the author's answer, shouldn't he use constant speed instead of average speed?

 

[PeroK]

Yes. According to Work-Energy theorem Work depends on change in speed.

Not if the mass starts and ends at rest, which is an unstated assumption. The work done is the same. Average speed gives the same answer as constant speed in this case.

 

[MatinSAR]

Not if the mass starts and ends at rest, which is an unstated assumption. The work done is the same. Average speed gives the same answer as constant speed in this case.

Thank you.
So according to your answer, The speed does not need to be constant. The how F is equal to mg? The object has acceleration ...

 

[MatinSAR]

I guess I understand something ...

You mean that because the total work is 0 and work of mg is constant we can calculate work of force F even if it is changing.

 

[PeroK]

Thank you.
So according to your answer, The speed does not need to be constant. The how F is equal to mg? The object has acceleration ...

The solution uses average force. Or, you could argue that if the speed profile doesn't matter, you might as well choose the simplest profile, which is constant speed. Note, however, that there must be at least one acceleration and deceleration phase. But, you can make these arbitraily short and hence negligible.

 

[PeroK]

I guess I understand something ...

You mean that because the total work is 0 and work of mg is constant we can calculate work of force F even if it is changing.

The work done by the varying force is $mgh$ in any case.

 

[MatinSAR]

The work done by the varying force is $mgh$ in any case.

I wasn't able to look at it in this way ...
Thanks a lot for your help @PeroK !

 

[erobz]

Homework Statement: An electric elevator lifts a 20kg body 40m at an average speed of 2m/s.(The motion is upward.)
Average power of the elevator?
Relevant Equations: $P_{av}=Work/Time$

The answer in the book is 400w. It said that $P_{av}=Fv_{av}cos(F,v)=mgv_{av}cos(F,v)=400$w and F is upward force that is applied by elevator.
Should velocity of the elevator be constant? Because it said that we have F=mg.

Just an engineering thought... they ask for "the average power of the elevator", maybe it's just me, but that roughly translates to "what is the output of the motor driving it". If I was a betting man I would say the weight of the elevator+ load is significantly more than the $20 \rm{kg}$ body it is lifting.

 

[Steve4Physics]

Just an engineering thought... they ask for "the average power of the elevator", maybe it's just me, but that roughly translates to "what is the output of the motor driving it". If I was a betting man I would say the weight of the elevator+ load is significantly more than the $20 \rm{kg}$ body it is lifting.

Possibly the elevator (of mass M) has an attached counterweight (also of mass M) and they move (frictionlessly) in opposite directions.

 

[erobz]

Possibly the elevator (of mass M) has an attached counterweight (also of mass M) and they move (frictionlessly) in opposite directions.

I never said I was a good engineer...

 

[scottdave]

It's 20 kg - it could be lifted in a bucket tied to a rope - with another bucket as counterweight.

 

[erobz]

It's 20 kg - it could be lifted in a bucket tied to a rope - with another bucket as counterweight.

What if some good engineer comes along, throws 20 kg in the counterweight bucket and ruins the physics problem?

 

[jbriggs444]

What if some good engineer comes along, throws 20 kg in the counterweight bucket and ruins the physics problem?

It is still a valid physics problem. One should always be clear about where the system boundaries are drawn.

Here, the [intended] interface in question is between the floor of the elevator and the bottom of the 20 kg object. We can ask about the average power flowing through this interface without pondering mechanical inefficiencies in the cables or hydraulics or the presence of counterweights.

Power is not invariant over choice of reference frame. But here we are given velocity with respect to some unspecified reference frame (presmably the rest frame of the building). So that part is not a problem.

It has already been noted that starting and ending velocities should have been specified. Presumably the elevator was moving up from the ground floor to some unspecified upper story 40 meters above and was at rest both before and after the motion.

Given the mass of the payload, we should perhaps be amazed that the child did not press all of the buttons. Or maybe a mad bomber placed the device in the cab and then spent 20 seconds briskly walking away.