# 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 mv^{2}

Ep = mgh

**3. The attempt at a solution**

I tried to assume that v_{i} was zero. Was that incorrect?

1/2 mv_{f}^{2} + mgh = 0

2.5v_{f}^{2} = -(5)(9.8)(10/(sin(20)))

v_{f}^{2}= -25.89130…

∴ Ek_{f} = 1/2mv_{f}^{2} = 64.7 N and ΔEk = 64.7 N

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

http://ift.tt/1eRWtYx

## Leave a comment