# Homework Help: Work, Power, Frictional Force - seems a bit tricky for me

1. Mar 4, 2012

### iJamJL

1. The problem statement, all variables and given/known data
A highway goes up a hill, rising at a constant rate of 1.00 m for every 48 m along the road. A truck climbs this hill at constant speed vup = 19 m/s, against a resisting force (friction) f equal to 1/24 of the weight of the truck. Now the truck comes down the same hill, using the same power as it did going up. Find vdown, the constant speed with which the truck comes down the hill.
ASSUME: the resisting force (friction) has the same magnitude going up as going down.

2. Relevant equations
W=F*s
F=μ*N

3. The attempt at a solution
Not really sure how to start this problem. What I've done so far is I've drawn this picture out, and that's all I can do. I drew it on a graph, and so the slope is 1/48 because the truck goes up 1 meter for every 48 meters it moves to the right of the origin. Then, the frictional force is applied downhill when the truck is traveling up, and the opposite when the truck is traveling down. I'm trying to find the power, or at least work, of the truck traveling up the hill, but it's not very clear because it doesn't say how far it traveled. We could use the measure of traveling to right 48m and 1m up, and then use that same amount to travel back down, but I'm still having trouble beginning this problem.

2. Mar 5, 2012

### tonit

$W = F * s$

$s = v * t$

since we have constant speed, this means the resultant force is 0.

Anyway, since when going down we have the same friction, the power remains the same too, what should the CONSTANT speed be? check it out and let us know

3. Mar 5, 2012

### iJamJL

Based on what you're saying, it sounds like it goes down the hill at the same speed (19 m/s) as it does going up. However, since we're going downhill, wouldn't we need to account for another factor? I'm not sure what exactly that is since:

mgsinθ - Ffriction = ma = 0
mgsinθ = Ffriction

Like you said, the resultant is 0 because the truck is traveling at constant velocity, and there is no acceleration. I assume I'm not seeing a key factor..

4. Mar 5, 2012

### iJamJL

Okay, so I realized that since this is an incline, we need to include it with the forces:

Going up,

F(truck) - Friction - mgsin(theta) = 0

That means going down,

-(F(truck) + mgsin(theta)) - friction = 0

I don't know whether to add them or set them equal to each other..

5. Mar 5, 2012

### LawrenceC

Hint:

Since the problem broaches the subject of power, equate power consumption going up to the power consumption going down. Power is force times velocity.

Last edited: Mar 5, 2012
6. Mar 6, 2012

### tonit

..................

Last edited: Mar 6, 2012
7. Mar 6, 2012

### LawrenceC

Hint: Let's keep it simple

Force to go up = W*sin(theta) + W/24
Force to go down = W/24 - W*sin(theta)

Power = force*velocity

You know the velocity up.

Equate power up to power down.