Kamden Larson

## Answered question

2022-10-13

I am doing an experiment where I extended a spring from a equilibrium a set distance and then launch a ball with it. Then I see how high it goes and what its average speed is (total distance it flies divided by time).
The machine that has the spring in it is set at an angle of 45 degrees. The aim of the experiment is to determine a relationship between distance from equilibrium of the spring to the height and velocity of the balls motion (2 different relationships).
The formula that express what happens could be given as, $0.5k{x}^{2}=mgh+0.5m{v}^{2}.$
My question is how much of the spring potential energy will be converted into gravitational energy and how much will be converted into kinetic energy (both at the highest point of the motion).

### Answer & Explanation

Laci Conrad

Beginner2022-10-14Added 17 answers

Let us ignore air resistance and set the launch position to be when gravitational potential energy is zero, i.e we assume it is launched from ground level. Therefore the total energy is just the initial kinetic energy, $\frac{m{v}_{0}^{2}}{2}$. Now at the maximum height, the vertical component of the velocity is zero, and hence the kinetic energy is $\frac{m\left({v}_{0}cos\left(\frac{\pi }{4}\right){\right)}^{2}}{2}=\frac{m{v}_{0}^{2}}{4}$ which we can see is half of the initial kinetic energy, therefore your intuition is right for this scenario.

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