# Max power in acceleration

1. Nov 14, 2009

### mike_302

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 degrees above the horiz. The car accelerates uniformly to a speed of 2.2 m/s in 12.0 s and then continues at a constant speed. What max power must hte winch provide?

2. Relevant equations

P=W/(delta)t W=(delta)E

3. The attempt at a solution

So I found the displacement using kinematics to determine the DISTANCE up the ramp that the cart goes while accelerating. this was 13.176m . Then I found the change in height by trig (6.588 m) And finally , I found the change in energy and divided by time of acceleration. My final answer was 5307 W . But the answer is supposed to be twice this value.

Where did we go wrong?

2. Nov 14, 2009

### Delphi51

The trouble is the power increases as the speed increases (P = F*v). In fact, the power is zero initially because the velocity is zero. The power at the end of the 12 seconds would likely be twice the average power you found.

I used a different method (Power = F*v) and got 10624 W.

3. Nov 14, 2009

### rl.bhat

Find acceleration due to gravity down the inclined plane. If you want to move the car up the shaft, first you have to overcome this retardation. Then accelerate the car to reach final velocity in 12 s. Maximum power = Net force*maximum velocity.

4. Nov 14, 2009

### mike_302

OHhhhhhh okay. We didn't cover that in our lectures so I have a feeling this question was just out there... to make us think. It's good though.

So let me get this straight: Power=Work/Time (constant power) Power=F*v (changing power)

5. Nov 14, 2009

### Delphi51

You can understand it this way:
P = Work/time = F*d/t
Since v = d/t, this is P = F*v
(Strictly speaking the d and t should be infinitesimally small to be correct when velocity is not constant.)

For the force, rl.bhat's suggestion is to write
sum of forces = ma
for the car. Both the forces and the acceleration are along the ramp so take only the component of mg that is along the ramp and ignore the component into the ramp.

6. Nov 15, 2009

### mike_302

Sorry... I think I need that explained a little differently. Which scenario is it better to use which of those equations for? I mean, I understand the derivation of those equations, but why use F*v rather than W/t in any given situation?

7. Nov 15, 2009

### rootX

Assume that you stop halfway. Find power spent in going from bottom to the middle point and then from the middle point to the top. Further find how power formula W/t was determined paying special attention to the assumptions made. To be honest, I should be looking at the derivation too. I never really cared much about the assumptions during high school/first year physics course or encountered this particular problem. Thanks for bring up this problem.

In this problem, rl.bhat has provided good explanation. Looking at the question again it only asks for instantaneous maximum power not average power over the whole trip.

Last edited: Nov 15, 2009
8. Nov 15, 2009

### Delphi51

Yes, it is better to use F*v than W/t because you need the instantaneous value at the end of the acceleration phase, not an average over a time.

9. Nov 15, 2009

### mike_302

Okayyyy. So either one of those equations works just as well for non accelerating systems, but the one with speed will find my maximum work given a specific speed. I understand. Thanks guys! That actually helped a tonne!