A Hooke's law spring is compressed 24 cm from equilibrium and the potential energy stored is 72 J. The spring is then used to launch a mass of 20 grams along a frictionless surface. The "launch speed" of the mass is?

Question

anonymous · Answer

The potential energy will be converted into kinetic energy $E_{k}=\frac{1}{2}mv^{2}$, from which you can calculate the velocity.

anonymous · Answer

Since energy is conserved: $$\Delta PE = \Delta KE$$ 
Therefore, if 72 J of PE is used to launch a mass, then the mass will have a KE of 72 J.
To find the velocity, use the following formula:$$KE = 0.5mv^2$$
Since you know KE and m, you can easily find v.

anonymous · Answer

Thanks guys, that's what I did, but I was just a bit hesitant when I evaluate the answer (84.85 m/s) because it just seems to fast. I was looking for a confirmation that that's really how you answer the problem.

anonymous · Answer

too fast*