Question

a vertical spring (ignore its mass), whose spring stiffness constant is 950 N/m, is attached to a table and is compressed down 0.150m a) what upward speed can it give to a .30 g ball when released? how high above its original position (spring compressed) will the ball fly?

Answer 1

First we need to calculate the potential energy of the compressed spring, right?

Answer 2

yes,,,then?

Answer 3

Well, what is that then and energy would be conserved/transferred from PotentialEnergy(spring) to KineticEnergy(ball), correct?

Answer 4

im referring to the procedure,,,
ahh..yes,,,your right,,,whats next?

Answer 5

Once you have the expressions for U(spring), K(ball) and relate them, you should be able to determine the initial velocity of the ball. Then it's a matter of pulling out your kinematic equations and you should be able to setup an equality relating the initial velocity to the displacement of the ball.

Answer 6

so,,,how will i answer it..?

Answer 7

Well, first of all, what are your expressions for potential and kinetic energy in this problem?

Answer 8

mg = kx, is this right?

Answer 9

That's Hooke's Law, which is related. The potential energy of a spring, U(spring) is:
\[U = \frac{1}{2}kx^2\]
What will be the expression for the kinetic energy of the ball?

Answer 10

1/2 mv^2?

Answer 11

Yes, then we have to relate them to each other, i.e. the form of the energy when the spring is compressed is equal to the value of the form of the energy after we let the ball go. Then if we solve for v, we should have the initial velocity of the ball.

Answer 12

i dont get it...pardon

Answer 13

To solve this we need to equate initial energy to final energy, in this case we have:
\[U(spring)=K(ball)\]or
\[\frac{1}{2}kx^2 = \frac{1}{2}mv^2\]
We can then solve for v, which will be the initial velocity of the ball.

Answer 14

ahhh !! the V is 21.79449473 m/s ,,,right?

Answer 15

I got around 267 m/s

Answer 16

how? what did you do?

Answer 17

\[\left( \frac{1}{2}kx^2=\frac{1}{2}mv^2 \right) * \frac{2}{1}\]
\[\left(kx^2 = mv^2\right)*\frac{1}m\]
\[\sqrt{\left( \frac{kx^2}{m}=v^2 \right)}\]
\[v = \sqrt{\frac{k}{m}}*x\]\[m = 0.30g * \frac{kg}{1000g} =0.00030kg=3.0kgtimes 10^{-4}\]
\[v = \sqrt{\frac{950 \frac{N}{m}}{0.00030kg}} * 0.150m = 26.9269563 \approx 267 m/s\]
Double checking my units I get.
\[\frac{m}{s}=\sqrt{\frac{\frac{N}{m}}{kg}} *m= \sqrt{\frac{\frac{kg*\frac{m}{s^2}}{m}}{kg}} * m =\sqrt{\frac{1}{s^2}}*m = \frac{1}{s}*m = \frac{m}{s}\]

Answer 18

slight error in the previous answer, it's supposed to be 266.9269563 in the sixth equation.