# Homework Help: Work Energy Theorem of a sled

1. Dec 8, 2008

### Jim4592

I don't really have any clue on this problem...

1. The problem statement, all variables and given/known data
A sled with mass 8.00 kg moves in a straight line on a frictionless horizontal surface. At one point in its path, its speed is 4.00 m/s. after it has traveled 2.5m beyond this point its speed is 6.00 m/s. Use the work energy relation to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sleds motion.

2. Relevant equations
W = ∆K

3. The attempt at a solution

The only thing i can think of (which isn't the work energy theorem) is Work = 1/2 MVf2-1/2MVi2

W= 1/2*(8kg)*(6m/s)2-1/2(8kg)*(4m/s)^2
W = 80 J

then use Work = FS Cos(Θ)

80J = F*(2.5m)*Cos(180)
-32N = F

2. Dec 8, 2008

### turin

Why do you think this isn't the work energy theorem? What do you think is the work energy theorem?

Why is Θ=180?

3. Dec 8, 2008

### Jim4592

i'm still getting 80 J for that one calculation, i called theta 180 because it was moving in a straight line.

4. Dec 8, 2008

### turin

Oh yeah, that's right. It never hurts to redo a simple calc, though, right? Sorry.

The value of theta has nothing to do with a straight line, it has to do with the angle between the directions of the force and direction traveled. theta is not 180. If theta were 180, then the force would be in the opposite direction from the motion, and so the object would slow down.