How Far Does the Ice Cube Travel Up the Slope After Release?

[freak_boy186]

Homework Statement

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?

Homework Equations

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

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...

 

[aim1732]

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.

 

[freak_boy186]

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...

 

[benhou]

apparently I'm incorrect in my reasoning...

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.

 

[benhou]

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

 

[aim1732]

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.