Change in Kinetic Energy for a Sliding Box

1. The problem statement, all variables and given/known data
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)))
vf2= -25.89130…
∴ Ekf = 1/2mvf2 = 64.7 N and ΔEk = 64.7 N

The correct answer is supposed to be 232.4 J. Help?

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