Physics question regarding spring energy; help?

Question

Question: A vertical spring (ignore its mass), whose spring constant is 900 N/m is attached to a table and is compressed 0.150m.
A) What speed can it give to a .300kg ball when released?
(Answer: 8.03 m/s)
B) how high above its original (compressed) position will the ball fly?
(Answer: 3.44 m)

I have the answers, I just need to find out how to get there.

My main question is how to find the velocity without knowing the distance it will fly, and vice versa. But I think my formula might be wrong, I've written:
SE (spring energy) = KE (kinetic energy) + U (potential energy)

I failed tremendously doing this, so I tried it without potential and got close to the right answer, but not exact, so I think i'm missing something. Help??

Thanks! I just needed to make sure I wasn't wrong in the formulas.

Answer 1

Your formula is not wrong. Quite the contrary. Now, since the ball is subject to gravity, its speed will vary in time -and distance, or, rather, height. Ball speed will be maximum at the point where it loses contact with the spring --beyond that point, there's no way it can increase any more.

Likely, this point is near the equilibrium point of the spring, i.e., when the spring is neither compressed nor stretched. If spring mass is negligible, there will be a negligible tendency to stretching. Besides, if, in spite of this, some stretching still occurs, the spring will be subject to an elastic restoring force which will oppose stretching.

Thus, a reasonable assumption is that ball speed will reach its maximum value when the spring regains its original length. Spring energy is ½ kx². KE = , and U = mgh. Therefore,

A)
½ kx² = ½ mv² + mgh
kx² = mv² + 2mgh
v² = k/m · x² – 2gh
v = √ (k/m · x² – 2gh) = √ [ (900/0.3) × 0.15² – 2 × 9.8 × 0.15 ] = 8.035 m/s.

B)
This is easier, since at this height all spring energy has been converted to potential energy.

½ kx² = mgh
h = k/m · x²/2g = 900/0.3 × 0.15²/19.6 = 3.44 m.

Answer 2

The speed that you want is when it leaves the spring. The final kinetic energy and gravitation potential energy at this point will equal the orginal energy stored in the spring.

The formula that you started with is correct SE = KE + U.
The height that you use for U is the compression distance 0.15m not the total height it goes because you want the speed as it leaves the spring.

To find the height you can use SE = U where U is the gravitational potential energy at it's highest point.

My first answer was wrong. I didn't think about the change in gravitational potential energy from the the distance the ball moves up while still attached to the spring

Answer 3

u have to equate the forces and then equate the sping co efficient with the distance its compressed with the force it is exerting on the spring balance these forces and u will get the answers