A 5.0 kg box slides up a 10 m long frictionless incline at an angle of 20 degrees with the horizontal, pushed by a 40 N force parallel to the incline. What is the change in kinetic energy?
2. Relevant equations
Ek = 1/2 mv2
Ep = mgh
3. The attempt at a solution
I tried to assume that vi was zero. Was that incorrect?
1/2 mvf2 + mgh = 0
2.5vf2 = -(5)(9.8)(10/(sin(20)))
∴ Ekf = 1/2mvf2 = 64.7 N and ΔEk = 64.7 N
The correct answer is supposed to be 232.4 J. Help?