# Distance travelled by the case down a ramp w/a spring at the bottom

1. Jun 7, 2009

### brunettegurl

1. The problem statement, all variables and given/known data

A 5.0 kg case of bottled Spring water is released from rest down a shipping ramp inclined 25oto the horizontal. At the base of the ramp, oriented parallel to its surface, is a spring that can be compressed 2.0 cm by a force 260 N. The case of water moves down the ramp and compresses the sring by 5.5 cm. How far has the case of water traveled down the ramp from its point of release before hitting the spring?
2. Relevant equations
Fspring= kx
E=mgh
E=0.5*kx2

3. The attempt at a solution
know that the force exerted by the spring is 260 at x=0.02 m i figured out k which is 13000N/m. i know that h= dcos $$\vartheta$$. i set up the equation
mgdcos$$\vartheta$$= 0.5*kx2 where x now equals the difference between 5.5cm and 2.0cm and solved for d i ended up getting 1.8m when the answer is actually half that i was wondering what i did wrong??

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2. Jun 7, 2009

### Noein

Check your expression for "h."

3. Jun 7, 2009

### brunettegurl

if i use sin theta i get -1.2278 m

4. Jun 7, 2009

### dx

It should be sin, not cos. And x is not the difference between 5.5 cm and 2.0 cm, it's just 5.5 cm, i.e. the amount by which the spring is compressed.

5. Jun 8, 2009

### brunettegurl

ok...so then the 2.0 cm was just given to us to figure out the force and has nothing to do with the actual compression of the spring when the bottle hits it..??

6. Jun 8, 2009

### Noein

Right. The 2.0 cm figure is only used to determine the spring constant.