Two masses are connected by a light string running over a frictionless pulley as shown below. The system is initially at rest and the incline is frictionless. If m1 = 10.0 kg, m2 = 8.00 kg and the incline makes a 30o angle from the horizontal, use energy methods to find the speed of the 8.00 kg mass just as the 10.0 kg mass reaches the ground 4.00 m below. (Hint: Think about how far the second mass will rise vertically as the first mass drops 4.00 m downward.)
KEf + Uf = KEi + Ui
(.5)mvf² + mgyf = (.5)mvi² + mgyi
The Attempt at a Solution
I'm really not sure what to do here. Here is one of my attempts.
(.5)(18)(vf)² +(18)(9.8)(0) = (.5)(18)(0)² + (18)(9.8)(4)
Solving for vf I get vf = 8.854, which is not the correct answer.
I'm pretty confused here and am not even sure if I am using the equation I need to be using. Any help would be greatly appreciated.