# Homework Help: Power and Energy

1. Aug 7, 2015

### Zynoakib

1. The problem statement, all variables and given/known data
A loaded ore car has a mass of 950 kg and rolls on rails
with negligible friction. It starts from rest and is pulled
up a mine shaft by a cable connected to a winch. The
shaft is inclined at 30.08 above the horizontal. The car
accelerates uniformly to a speed of 2.20 m/s in 12.0 s
and then continues at constant speed. (a) What power
must the winch motor provide when the car is moving
at constant speed? (b) What maximum power must
the winch motor provide? (c) What total energy has
transferred out of the motor by work by the time the
car moves off the end of the track, which is of length
1 250 m?

2. Relevant equations

3. The attempt at a solution
I actually can get the answer but I just don't understand th

The way I calculate it is that add part 2 and part 3 together and divide them by 12 and then add part 1 into the answer.
I understand why word done by acceleration have to be divided by 12 (to convert it to power), but I just don't understand why I need to do the same thing for KE, if I divide it by 12, then it will become the power required to accelerate the object from 0 m/s to 2.2 m/s, which is exactly the same thing as having the word done by acceleration divided by 12. So, why do I need to divide KE by 12? Or the entire thing is just plain wrong?

Thanks!

Last edited: Aug 7, 2015
2. Aug 7, 2015

### billy_joule

I can't quite understand what you're saying. What are part 1, 2, &3? And your handwriting is illegible.

From what I can tell you've worked out average power over the 12 second acceleration stage, this is not the maximum power.
Max power occurs only for a single instant in this case.

Are you familiar with P=Fv?

3. Aug 7, 2015

### haruspex

Nowhere in your working do you mention the work done hauling the car up the gradient.
The very first line quotes the power at constant speed, suggesting you have figured this out somewhere else.
Along with billy_joule, I cannot make snse of your references to parts 1 to 3. They don't seem to equate with a to c since you describe adding 2 and 3 together, but a and b answers are power, while c is work.

4. Aug 8, 2015

### Zynoakib

I am sorry for causing so many confusions by the way I ask and my handwriting.

Thanks for your P=Fv, I really rang a bell to me and I have solved the problem.

But right now I have another question:

My original way of calculating (b) was to find the force of acceleration of the block first, whichi is
u = 0, a = ?, v = 2.2, t = 12
v = u + at
2.2 = 12a
a = 0.18333 ms^-2

F= ma = (950)(0.1833) = 174 N

Then, I found the the distance traveled by the block
u = 0, v = 2.2, s = ?, a = 0.18333
v^2 = u^2 + 2as
2.2^2 = 2(0.1833)s
s = 13.2 m

Then I found the work done by acceleration
W = Fs = (174)(13.2) = 2296.8

Power = 2296.8/ 12 = 191.4 W

Total power = 1024 (calculated from (a)) + 191.4 = 1215 W, which is wrong.

So why can I not obtain the right answer with such approach?

5. Aug 8, 2015

### haruspex

You have taken the total work done in the acceleration phase and divided by the time taken. That will give you the average power. You are asked for peak power. Since the force is constant but the velocity is increasing, the power, Fv, increases during this phase.
Not sure what your thinking is there. Is your idea that the total power consists of the power which provides acceleration plus power which maintains the speed? It doesn't work like that. If you apply a force F to an object while that object is currently moving at speed v in the direction of the force then the power you provide is Fv, regardless of whether the object is accelerating. Whether it accelerates will depend on how F compares with the other forces.

6. Aug 8, 2015

### billy_joule

In addition to haruspex comments it appears your work is for the level track case, that is, you've ignored the incline.