When the spring is compressed, the potential energy stored in the spring is:
[tex]U= \frac{1}{2}kx^2 [/tex]
where k=880 N/m is the spring constant and [tex]x=0.170 m[/tex] is the compression of the spring. By using these numbers, we get
[tex]U= \frac{1}{2}(880 N/m)(0.170 m)^2=12.7 J [/tex]
When the spring is released with the ball over it, all the potential energy is converted into kinetic energy of the ball:
[tex]U=K= \frac{1}{2} mv^2[/tex]
So by using m=0.300 kg, and re-arranging the formula, we can calculate the velocity of the ball:
[tex]v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 12.7 J}{0.300 kg} }=9.2 m/s [/tex]
