Conservation of energy, speed before case hits spring

In summary: U_g = K \\mgh = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \\12.9(9.81)(0.406\sin{27.2}) = \frac{1}{2}(12.9)v^2 + \frac{1}{2}(\frac{280}{2.12})(0.0596)^2 \\v = \sqrt{\frac{12.9(9.81)(0.406\sin{27.2}) - \frac{1}{2}(\frac{280}{2.12})(0.0596)^2}{\frac{1}{2}(12.9)}}
  • #1

Homework Statement

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?

Homework Equations

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?
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  • #2
Is it 1.65m/s?
  • #3
grzz said:
Is it 1.65m/s?

Nope. :(
  • #4
Sorry i meant to say 1.87m/s
  • #5
grzz said:
Sorry i meant to say 1.87m/s

Not that either... :(
  • #6
First use the information to solve for the constant of the spring:
F = kx \\
280 = (2.12)k \Rightarrow k = \frac{280}{2.12}
Then use the work energy theorem to solve for the velocity of the water as it came in contact with the spring by setting its kinetic energy equal to the difference in the work done by gravity and by the spring in terms of the compression of the spring.

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