The speed of the mass v = 0.884 m/s.
Explanation:
Let
K1 represents the kinetic energy of the mass when it is released,
U1 represents the potential energy of the spring when the mass is released,
K2 represents the kinetic energy of the mass when the spring returns to relaxed length,
U2 represents the potential energy of the spring when the spring returns to relaxed length
The spring is stretched by 0.27 - 0.12 = 0.15 m
K1 = 0
U1 = (1/2) [tex]\times[/tex] 0.8 [tex]\times[/tex] (0.15)^2
= 0.009 J
U2 = 0
By conservation of energy,
K2 + U2 = K1 + U1
K2 + 0 = 0 + 0.009 J
K2 = 0.009 J
Let v = speed of the mass
K2 = 1/2 [tex]\times[/tex] m [tex]\times[/tex] v^2
m = 23 g = 0.023 kg
0.009 = 1/2 [tex]\times[/tex] 0.023 [tex]\times[/tex] v^2
0.009 = 0.0115 [tex]\times[/tex] v^2
v = √(0.009 / 0.0115)
v = 0.884 m/s.
