In our physics lab we fired our spring loaded launcher using

In our physics lab we fired our spring loaded launcher using the yellow ball which had mass of 9.8 grams. If the spring is compressed 19.91 cm and the ball travels horizontally 2.15 m from a height of 0.79 m, determine the spring constant of the spring.

Solution

The energy stored in the spring is going to equal the kinetic energy produced when the spring is released.

0.5kx^2 = 0.5mv^2
k = mv^2/x^2 = (0.0098kg)v^2/(0.1991m)^2 = 0.2472*v^2

How do you find v one might ask? In this case we know that the spring shoots the ball horizontally, so all of the kinetic energy produced is horizontal. Therefore the v we are looking for is horizontal only.

We know that the ball travels 2.15 m horizontally in the same amount of time that it free falls to the ground .79m below. We first find the amount of time this takes using the kinematic equation.

d=v0*t + 0.5at^2
.79 = 0 + .5(9.8)t^2
t = .401 s

Now we know it takes .401s for the ball to hit the ground.

We also knows that once the ball starts to travel horizontally, it does no longer accelerates horizontally, so we use the simple equation.
d = v*t
v = d/t = 2.15m/.401s = 5.361 m/s

Now we know the initial velocity, so we can solve the equation we set up earlier for k.

k = 0.2472*v^2 =

k = 7.104 N/m