A 12.9 kg case of bottled water is released from rest down a shipping ramp inclined 27.2° to the horizontal. At the base of the ramp, oriented parallel to its surface, is a spring that can be compressed 2.12 cm by a force of 280 N. The case of water moves down the ramp and compresses the spring by 5.96 cm. At what speed is the case moving just as it touches the spring?
Ug = K
The Attempt at a Solution
I found the distance that the case travels to be 0.406m. I know this is correct.
Ug = Ek
0.406sin(27.2) * 9.81 = 1/2(v^2)
v = 1.91 m/s
This is wrong. what am I doing wrong?