Velocity and Acceleration at Maximum Compression (help, test tomorrow?)

[HeyBurrito]

Velocity and Acceleration at Maximum Compression (help, test tomorrow??)!

Homework Statement

A 300g ball compresses a spring 1 cm when just sitting on it (no motion). The ball is dropped onto the spring from 25 cm above the spring at equilibrium. How much will the spring compress as it stops the ball? At that maximum compression, what is the velocity and acceleration of the ball?

Homework Equations

W= 1/2(m)(delta X)^2
PE= mgh

The Attempt at a Solution

I got W=.0015 from 1/2(3)(.01)^2
and .7358 PE from .3(9.81)(.25)
I can't find the velocity, but acceleration I attemped F=ma and got stuck there lol.

 

[irycio]

Sorry, I don't really catch your attempt at a solution, so here is mine :P.

Information you are given in a first sequence allows you to calculate the spring constant, usually denominated with k (F=kx, where x is the change of length).

Now, you need to choose the best zero-level for the potential energy of your ball. My suggestion is to choose it not at the equilibrium, but at the very point, that the falling ball will compress the spring to, so that the whole potential energy of the ball is transformed into the energy of the spring, given as (1/2)k*x^2. It may seem more complicated, but believe me, it isn't and gives you a better understanding of what is happening as well.
And now the last question. The question about the velocity you have already answered in your question-"as the ball STOPS". What do you think is the velocity of the stopped ball??:P
Now, the acceleration of the ball is F/m, obviously, F being the force the spring acts on the ball with. And this very force depends on the compression, as already mentioned.
Hope it helped

 

[HeyBurrito]

Hey! You did help! Vf=0 haha I am so silly. I'm going to rework it now. (:

 

[HeyBurrito]

so when doing 1/2(30kg)(.01m)^2= .0015So, PE= (.3)(9.81)(.25+.1)
PE=1.03

1.03=1/2(.3)(V^2)
V=2.62 M.s

2.62^2=0+2(a)(.1m)
6.86=2(a)(.1)
a=34.3 m/s^2?

 

[HeyBurrito]

oh and i didn't understand the bit about force because i got something cooky. Gah, I fail at Physics.

 

[irycio]

300g=0.3kg, to start with.
But let's use symbols only, hence F being force, m mass of the ball, k-spring constant, x1-first compression, x2-2nd. compression.

Start with the equality of the values of two forces-ball's "m*g", g=9,81 m/s^2, and kx1; this is according to Newton's 3rd rule of dynamics.
You get the equation:
mg=kx,
which let's you calculate the spring constant k.

Now, your calculation pf PE is incorrect, as you assumed, that the 2nd compression wil be 1 cm (0.01 m) as well, which is incorrect, as it will be larger :P.
Start with the potential energy
PE=mgh
where h is the height of the ball above zero level. As i already mentioned, i recommend zero-level to be at the end of the compressed spring after the ball stops. Hence h=0.25+x2
And now this potential energy is transformed into spring's energy, given by the equation
E=(1/2)*k*(x2)^2.
The equality of potential energy of the ball and spring's energy is a quadratic equation for x2. Solve it, choose the proper solution (x2 has to be larger than 0). And now you are able to calculate the force, the spring acts on the ball with, which will lead you to an acceleration.