# Homework Help: Conceptual Question about Power

1. Feb 27, 2012

### PeachBanana

1. The problem statement, all variables and given/known data

There are two paths up a hill, a steep one and a
less steep one. If we travel at the same speed
along the paths, what can we say about your
power in climbing the hill?

A) more on the steep one
B) more on the less steep one
C) the same on each
D) it depends on how much time it takes you
travel over path

2. Relevant equations

P = F * v * cos (theta)

3. The attempt at a solution

I've eliminated C and D because the velocity is the same for both of them. It doesn't depend upon how much time it takes to travel over the path because that's the same. The angle between the force and the displacement is zero but the angles of the hills aren't (steep one will have an angle much bigger than the less steep one) but I'm not sure how to incorporate that into the above equation.

Last edited: Feb 27, 2012
2. Feb 27, 2012

### tiny-tim

Hi PeachBanana!
how can the speed be the same if the distances are different?

and how do you know what the forces are?

you may find it easier to use power = work done per time

3. Feb 27, 2012

### LawrenceC

Hint: Draw two right triangles, one a 45,45,90 and the other a 30,60,90. Let both have the same altitude. Compute work done per unit time to find your answer. You can determine time needed for each path (triangle hypotenuse) by some simple calculations.

4. Feb 27, 2012

### PeachBanana

I'm going to go ahead and say "A" because a steeper hill will have a larger angle.

W = F * d * cos (theta)
W = mg * d * cos (theta)

Power = W / t. I'm assuming t is the same but the path up the steeper hill will do more work.

5. Feb 27, 2012

### John61

if you travel the same distance at the same speed you need more power for the steeper path, of course you also go higher as the angle is greater, or have to travel further on the less steep path to reach the same place

6. Feb 28, 2012

### LawrenceC

Wait a minute here. The force to push the weight up the hill is

mg * sin(theta)

Your force is the normal force. No work done there because it is at right angle to force. Secondly, t is not the same.

The easiest way to reason this out is to recognize that no matter what path you take, the work done is mgh. Power is work divided by the time it took to do it. So if the velocity is the same as stipulated in the problem, the time is smallest for the shortest path which is straight up. The less the slope, the longer the path is so therefore the time is greater. The more time it takes to do the work, the less power is expended.

7. Feb 28, 2012

### PeachBanana

@LawrenceC thank you. That was very helpful.