# Power question

1. Dec 9, 2015

### imsmooth

My daughter showed me a question:
Car A and B have equal mass and go up a hill. A has twice the constant velocity of B. Compared to B the power of A is?

I found these answers here which seem to miss the point:

However, the KE is 4x greater for car A than B.
Also, the time for car A to go up is 1/2 that of B because of the doubled velocity.

Since Power is work/time shouldn't the power of A be 8x greater than B?

2. Dec 9, 2015

### Staff: Mentor

No, I'm not a fan of the wording, but since KE isn't changing, you can't get a power from it. So the problem can only be referring to the change in potential energy over time.

Last edited: Dec 9, 2015
3. Dec 9, 2015

### nasu

You don't need to go through all that KE and work steps.
Power can be also written as
P=F*v (Force times velocity).
As they move with constant velocity, the force should balance the tangential gravity which is the same for both (they have same mass).
So same force, twice the velocity means twice the power.

4. Dec 10, 2015

### CWatters

The problem doesn't say the car starts from rest so you have to assume they start with KE and 4xKE. That extra speed and KE could have come from the engine some time earlier but the problem is only concerned about the power needed to go up the hill. The initial higher KE could also have come from a previous downward slope. The point is that the KE isn't changing as you go up the hill so it makes no difference to the power required.

That only reduces how long the extra power has to be delivered for not the amount of power.

If you were asked to calculate the Energy required to get to the top that would be the same in both cases but you might calculate them differently....

Slow car... Energy = Power * Time
Fast car... Energy = 2*Power*Time/2

The fact that both cars use the same energy shouldn't be a surprise because the hill is the same height in both cases. In the real world the faster car would consume more due to increased drag but that wasn't mentioned.