Question

Trying to figure this one out. A vertical spring whose spring constant is 875N/m is attached to a table and is compressed down by .160m. What upward speed can it give to a .350kg ball when released? How high up its original position will the ball fly?

Answer 1

I think can figure out the second part of the question, I'm just confused on the first part.

Answer 2

I know I'll need the conservation of energy equation, and I also know that at point 1 potential energy is created by the compression of the ball on the spring and at point two kinetic energy is created when the ball is released from the spring and shot up straight in the air.

Answer 3

So the relevant equation I'll need are \[1/2kx^2\]
Sense the only force acting on the spring at point 1 is the force of gravity,\[1/2mv _{1}^{2}+mgy _{1}\] and at point two the spring is moving along with the ball, so wouldmy equation look something like this:
\[1/2mv _{1}+mgy _{1}=1/2mv _{2}+1/2kx^2\]

Answer 4

If you consider the initial position of the spring to be y=0, all the potential energy of the spring is utilized to give the object a velocity.
kx^2/2=mv^2/2
Then all the kinetic energy of the body is then converted into its potential energy. At the highest point, kinetic energy is 0.

Answer 5

1/2kx2=mgh