To solve this problem it is necessary to apply the concepts related to the conservation of Energy. In this case we must apply the conservation of the initial kinetic energy, as well as the final kinetic energy together with the elastic potential energy. Mathematically this can be expressed as,
[tex]KE_i = KE_f + PE_f[/tex]
[tex]\frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2+\frac{1}{2}kx^2[/tex]
Where,
m = Mass
[tex]v_i[/tex]= Initial Velocity
[tex]v_f[/tex]= Final Velocity
k = Spring constant
x = Displacement
Our values are given as,
[tex]m = 2 Kg\\k = 800 N/m\\x = 10 cm = 0.10 m\\v_i = 4 m/s[/tex]
Replacing we will have to,
[tex]\frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2+\frac{1}{2}kx^2[/tex]
[tex]\frac{1}{2}(2)(4)^2 = \frac{1}{2}(2)v_f^2+\frac{1}{2}(800)(0.1)^2[/tex]
Solving to find the final velocity we have that,
[tex]v_f = 3.4641m/s[/tex]
Therefore the speed when it is a distance of 10cm from the equlibrium position is 3.4641m/s
