How Does the Work-Energy Theorem Apply to Calculating Force on a Sled?

[Jim4592]

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

Homework Statement

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.

Homework Equations

W = ∆K

The Attempt at a Solution

The only thing i can think of (which isn't the work energy theorem) is Work = 1/2 MV_f²-1/2MV_i²

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

 

[turin]

The only thing i can think of (which isn't the work energy theorem) is Work = 1/2 MV_f²-1/2MV_i²

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

Work = 1/2 MV_f²-1/2MV_i²
W = 80 J

Redo your calculation here.

8.00 kg
moves in a straight line on a ... horizontal surface
At one point ... its speed is 4.00 m/s
2.5m beyond this point its speed is 6.00 m/s
find the force acting on the sled
assuming ... this force is constant and that it acts in the direction of the sleds motion

Work = FS Cos(Θ)

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

Why is Θ=180?

 

[Jim4592]

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

 

[turin]

i'm still getting 80 J for that one calculation,

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

i called theta 180 because it was moving in a straight line.

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.