Answer:
(D) the speed of the mass when the spring returns to its natural length is 1.58 m/s.
Explanation:
Given;
spring constant of the spring, k = 2 N/m
mass attached to the spring, m = 0.2 kg
compression of the spring, x = 0.5 m
Apply the principle of conservation of mechanical energy;
K.E = P.E
¹/₂m(v² - u²) = ¹/₂kx²
where;
u is the initial speed of the mass = 0
¹/₂mv² = ¹/₂kx²
mv² = kx²
[tex]v^2= \frac{kx^2}{m} \\\\v= \sqrt{\frac{kx^2}{m}} \\\\v = \sqrt{\frac{(2)(0.5)^2}{0.2}} \\\\v = 1.58 \ m/s[/tex]
Therefore, the speed of the mass when the spring returns to its natural length is 1.58 m/s.
