# Ball falling on Spring.

1. Dec 5, 2013

### Patolord

1. The problem statement, all variables and given/known data
A sphere of 4kg falls from a height of 70cm and lands on a spring of constant 2 *10² N/m.
Whats the max velocity the sphere reachs?
g=10m/s²

2. Relevant equations
Ep=m*G*h
Ek=mv²/2
Eel=k*x²/2
Vmax=sqrt(2g*h) (internet source idk if this is right)

3. The attempt at a solution
I tried this Ep = Eel + epx
Ep=4*0,7*10
28=kx²/2 + mg*x
x=0,367 m
them i did 0,7 + 0,367 for the height
and it gave me Sqrt(2g*h)=v
V=4,62 m/s
not the answer, i couldnt get my head around the energies. I also tried Ep=Eel but no sucess

2. Dec 5, 2013

### Staff: Mentor

These are OK.

This only applies to a body falling freely from a height h. Don't need it.

The total energy is (using your notation): Ep + Ek + Eel
That total energy is conserved.

Hint: At what point does the mass stop accelerating? (After that point, it starts slowing down.)

3. Dec 5, 2013

### Patolord

Sorry, power went out on my house.
So it stops acelerating when it touches the spring.
But i still need the energy of the spring right because it touches it?
If i try ep+ek + Eel i get 28 -2v² = 200x²/2
now to find the x
i can use the weight of the ball F=m*a = k *x , i get 40=200*x => x=0,2
them Eel = k*x²/2 = 4 !
Eel+ Ep = Ec Ec = 32
32= 4 * v² /2
16 = v²
v=4 m/s Yeaah lol solved while writing this thank you veryy much ! (or maybe i did something wrong and got to the right result ?)