Hooke's experiment and spring launcher

[starrish]

Homework Statement

Hooke's Law Experiment: what is force and extension table of values?, finding the spring constant and calculating one initial launch velocity of my spring launcher?

Homework Equations

Fx=k(x)

The Attempt at a Solution

researching on google...need someone to clear up the concept

 

[unscientific]

Ok, basically hooke's law states that within an elastic limit of a material, the Force is directly proportional to its extension, given by F = kx, where x is the extension in metres, and k is the spring constant with units [N/m].

Say, a spring has a spring constant of 360N/m, so for every 1m u compress the spring, the spring exerts a force in opposite direction of 360N.


Looking at the thumbnail attached, u can see that its a force-extension graph illustrating hooke's law, with gradient of k [360N/m]. To calculate the ELASTIC POTENTIAL ENERGY stored in the spring, its Force x distance, which is the area under the graph, shaded in blue.

Therefore, Elastic Energy = 0.5Fx = 0.5(kx)x = 0.5kx².

Assuming a spring is compressed by x metres, it contains elastic potential energy of 0.5kx². Then following the law of conservation of energy, where all elastic potential energy is converted into kinetic energy of the ball being launched.

0.5kx² = 0.5mv²
kx = mv²

v = sqrt(kx/m)

Hope this clarifies your doubt.

 

[unscientific]

force-extension diagram

This is the force extension diagram as described in my previous post.

 

[starrish]

thank you that helped a lot...and ur username is very ironic. But other than that there's a new problem for me now, how do i make an energy and distance/range graph. And also i have noticed that as the angle value for my launcher goes higher, the distance my spring travels increases. But since the higher angle created more height in the projectile motion created by my spring wouldn't my results be vise versa, and the higher the angle is...the lower the distance?