Answer:
1.75 m/s
Explanation:
k = Spring constant = 490 N/m
m = Mass of object = 5.8 kg
x = Displacement of spring = 0.19 m
v = Speed of object at the equilibrium position
The potential energy of the spring will balance the kinetic energy of the mass
[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{kx^2}{m}}\\\Rightarrow v=\sqrt{\dfrac{490\times 0.19^2}{5.8}}\\\Rightarrow v=1.74637\ \text{m/s}[/tex]
The speed of the mass as it returns to the equilibrium position is 1.75 m/s.
