Maximum power developed by a lift

[MNWO]

Hi guys, I've been given a question where it asks me to work out : work done , the tension in lifting cable and the maximum power developed and the problem is that I don't know what the maximum power developed means. Isn't it the same as the work done ? If its not, then what's the formula to work out the maximum power developed? Here is the question just in case ;)
A lift cage of mass 539kg accelerates upwards from rest to a velocity of 6m/s whilst traveling a distance of 14m. The frictional resistance to motion is 205 N. Making use of the principle of the conservation of energy, determine: work done , the tension in lifting cable , the maximum power developed.

 

[sophiecentaur]

Power is rate of doing work. But, of course, with a counterweight, during the actual climb, at uniform velocity, you would only need to be overcoming friction. Serious power is needed during the acceleration phase.

 

[NTW]

Another approach could be to assume that there's no counterweight. So, concerning work, you should remember two portions:

1) The work associated with the force of friction...
2) The increase in potential energy of the cage due to the climb...

And, concerning the tension in the lifting cable, I count three forces to be reckoned with... And to be calculated and added...

Although Sophiecentaur states that the climb is at uniform velocity, I understand, from the formulation, that the lift moves with a constant acceleration, so the power needed is uniform during the climb, and -knowing the acceleration and the distance- you could calculate the time needed. You know already the work done, and as power = work/time...

 

[sophiecentaur]

Another approach could be to assume that there's no counterweight. So, concerning work, you should remember two portions:

1) The work associated with the force of friction...
2) The increase in potential energy of the cage due to the climb...

And, concerning the tension in the lifting cable, I count three forces to be reckoned with... And to be calculated and added...

Although Sophiecentaur states that the climb is at uniform velocity, I understand, from the formulation, that the lift moves with a constant acceleration, so the power needed is uniform during the climb, and -knowing the acceleration and the distance- you could calculate the time needed. You know already the work done, and as power = work/time...

What would be the point of assuming no counterweight? What sort of engineer would miss out an all that free GPE every time the lift operates? But perhaps the question was written by a Physics Teacher. It would only need for the word "cage" to be changed into "load" and it would make sense for and Engineer and a Physicist.
It would be only the GPE of the load that is significant because the counterweight is chosen to be a bit heavier than the cage itself. But enough said.

As for the question. I reckon you just need to find the accelerating force needed and add it to the friction force. Then Power is Force times Speed. If the accelerating Force is constant and the speed is increasing, the maximum Power will be when it's going fastest.

 

[NTW]

Well, no counterweight is mentioned in the formulation. I know it's not a normal thing in lifts, but...

I have my solution to the problem, that I believe to be correct, but don't want to publish it now, until MNWO gives his...

 

[sophiecentaur]

Well, no counterweight is mentioned in the formulation. I know it's not a normal thing in lifts, but...

I have my solution to the problem, that I believe to be correct, but don't want to publish it now, until MNWO gives his...

You are right, of course, about the question itself. Science education is full of test questions that could be made much better with just a little extra thought.