Answer:
[tex]31.6\:\mathrm{m/s}[/tex]
Explanation:
The elastic potential energy of a spring is given by [tex]Us=\frac{1}{2}kx^2[/tex], where [tex]k[/tex] is the spring constant of the spring and [tex]x[/tex] is displacement from point of equilibrium.
When released, this potential energy will be converted into kinetic energy. Kinetic energy is given by [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the object's velocity.
Thus, we have:
[tex]Us=KE,\\\frac{1}{2}kx^2=\frac{1}{2}mv^2[/tex]
Substituting given values, we get:
[tex]\frac{1}{2}\cdot 50\cdot 0.2^2=\frac{1}{2}\cdot 0.002\cdot v^2,\\v^2=\frac{50\cdot 0.2^2}{0.002},\\v^2=1000,\\v\approx \boxed{31.6\:\mathrm{m/s}}[/tex]
