# Spring friction physics

1. Mar 11, 2010

### freak_boy186

1. The problem statement, all variables and given/known data
A 50 g ice cube can slide without friction up and down a 30 degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 10 cm. The spring constant is 25 N/m. When the ice cube is released, what total distance will it travel up the slope before reversing direction?

2. Relevant equations
F = ma
W = Fdx
Ws = -1/2kx^2

3. The attempt at a solution
Ws = -1/2(25)(.1)^2 = 0.125J
V1 = [2(.125/.5)]^1/2 = .7071 m/s
F = .5(9.8)(sin30) = 2.45N
2.45/.5 = 4.9 m/s^2

(.7071)^2 = 2(4.9)ds
ds = .0510m = 5.1cm

apparently I'm incorrect in my reasoning...

Last edited: Mar 11, 2010
2. Mar 11, 2010

### aim1732

Re: springs

When the spring decompresses there is change in K.E as well as P.E. You have considered only change in K.E. Rather look at the end positions that concern you most - the elastic P.E at maximum compression will be equal to G.P.E when block comes to rest momentarily at highest pt.

3. Mar 11, 2010

### freak_boy186

Re: springs

so what is the equation for P.E. then, cause I thought it was -1/2kx^2...

So... EPIC fail... 5g = .05kg

My answer WAS correct, the decimal place was just off by one...

Last edited: Mar 11, 2010
4. Mar 11, 2010

### benhou

Re: springs

I am not sure what you mean by that. But, your solution looks reasonable. There is an easier way to do it though. simply set $$U_{s}=U_{g}$$ without the kinetic energy.

5. Mar 11, 2010

### benhou

Re: springs

By the way, 50g=0.05kg, not 0.5kg.

6. Mar 11, 2010

### aim1732

Re: springs

Dear freak_boy I wasn't referring to elastic potential energy rather it was gravitational potential energy . That is why your answer was off by a small margin - the change of height you ignored was small. Nonetheless it is important that you do not ignore it. That is why benhou's method is better.