Calculating Speed & Height of a Spring-Released Ball

[chazgurl4life]

A vertical spring (ignore its mass), whose spring stiffness constant is 900 N/m, is attached to a table and is compressed down 0.110 m.
(a) What upward speed can it give to a 0.300 kg ball when released?
-----------m/s
(b) How high above its original position (spring compressed) will the ball fly?
------m


Now if I am following my lecture notes correctly for part a)
I am using the equation is that V^2=KX^2
-------
mass of ball
so when we out it all out it should be v^2=900n/m(-.110m)^2
-----------------
.300 kg

-------------------------->v=6.02 m/s but that answer is wron apparently. I don't know where I am going wrong.

 

[Hootenanny]

I wouldn't mind seeing the derivation for this formula, do you have it?

V^2=KX^2

 

[chazgurl4life]

the original equation is
.5mass1v^2+ .5kx^2=1/2mass2v^2+ .5kx^2
==>0+1/2mv^2=1/2Kx^2 +0
===>v^2=kx^2/mass

 

[Hootenanny]

the original equation is
.5mass1v^2+ .5kx^2=1/2mass2v^2+ .5kx^2
==>0+1/2mv^2=1/2Kx^2 +0
===>v^2=kx^2/mass

That look's better, however, you are still ignoring gravitational potential. Are you happy assuming the GPE is negligable?

 

[chazgurl4life]

so how do i intergrate that with this problem?

 

[Hootenanny]

What is the equation for GPE?

 

[chazgurl4life]

the equation that i was taught was that U=MGH

 

[Hootenanny]

well there you go, just add this in like so;

$$\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + mgh$$

However, if you haven't incorperated GPE in this type of porblem before, I would ask you tutor before doing so.

 

[chazgurl4life]

ok so since we are ignoring mass the:


.5(900N/M)(.110^2)=.5(v^2) +(h)(9.8)