Answer:
Speed, v = 6.91 m/s
Explanation:
Given that,
Spring constant, k = 594 N/m
It is attached to a table and is compressed down by 0.196 m, x = 0.196 m
We need to find the speed of the spring when it is released. Here, the elastic potential energy is balanced by the kinetic energy of the spring such that,
[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{kx^2}{m}}[/tex]
[tex]v=\sqrt{\dfrac{594\ N/m\times (0.196\ m)^2}{0.477\ kg}}[/tex]
v = 6.91 m/s
So, the speed of the ball is 6.91 m/s. Hence, this is the required solution.