# Homework Help: Power and energy

1. Jan 17, 2016

### Rumplestiltskin

1. The problem statement, all variables and given/known data
A 55kW motor is used to lift a 4800kg mass vertically up a mine shaft. What is the maximum possible speed that the mass could move upwards? Give your answer to 2 significant figures.

2. Relevant equations
3. The attempt at a solution

I correctly answered this as 55000 / (4800 * 9.8) = 1.2 m/s.
But first I went on a tangent rearranging KE = 0.5mv2 for v. Seemed like the most intuitive line of thought. Why was this mistaken? How would I avoid wasting time like that?

2. Jan 17, 2016

### haruspex

It's not going to work because there is no change in KE. The change is in PE.
That's intuitively obvious to me, but I have no magic formula to make it intuitive for you. Trying to analyse my own thought processes on the problem: it's a question about power, and power is the rate of work done; what work is being done?; it's the work in raising the mass against gravity, so it's the rate of change of PE....

3. Jan 17, 2016

### Rumplestiltskin

Why would there need to be a change? If you knew the KE and neglected friction, couldn't you work out the velocity?

4. Jan 17, 2016

### CWatters

+1

For this type of problem you need to decide which is changing, the PE, the KE or both.

5. Jan 17, 2016

### haruspex

How are you going to know the KE? You are not given KE, it is not changing, so the only way to know it is by knowing the velocity and mass, but the velocity is what you are trying to find.
I think this is the key intuition, that you need to look at what is changing.

6. Jan 17, 2016

### Rumplestiltskin

Would I be able to find if it were changing?

Last edited: Jan 17, 2016
7. Jan 17, 2016

### haruspex

If the KE were changing? In that case the velocity would be changing, so you'd need to specify the question as velocity at some particular stage in proceedings.

8. Jan 17, 2016

### Rumplestiltskin

Maybe not always; think a rocket burning fuel at constant velocity. But I see your point. I guess the intuition will come with practice.

9. Jan 17, 2016

### haruspex

But there you are treating the rocket and remaining fuel as the mass. That is not a consistent object, i.e. it is not the same object from one time to another. You would have to include the exhaust fuel, which has quite a different velocity.

10. Jan 18, 2016

### CWatters

The problem statement asks for the maximum velocity. There is only one maximum so the maximum velocity cant be changing. At any other velocity the power won't be 55kW.