Answer:
The speed of the ball when it reaches equilibrium position is 3.31 m/s
Explanation:
Given;
mass of the object, m = 0.25 kg
initial displacement of the object, h₁ = 0.56 m
spring constant, k = 105 N/m
displacement at equilibrium position, h₂ = 0
initial velocity of the object, v₁ = 0
velocity of the object at equilibrium position = v₂
The change in gravitational potential energy at the equilibrium position is given as;
ΔP.E = mg(h₂ - h₁)
The change in kinetic energy of the object at the equilibrium position is given as;
ΔK.E = ¹/₂m(v₂² - v₁²)
Apply the principle of conservation of mechanical energy;
ΔK.E + ΔP.E = 0
¹/₂m(v₂² - v₁²) + mg(h₂ - h₁) = 0
¹/₂m(v₂² - 0) + mg(0 - h₁) = 0
¹/₂mv₂² - mgh₁ = 0
¹/₂mv₂² = mgh
¹/₂v₂² = gh
v₂² = 2gh
v₂ = √2gh
v₂ = √(2 x 9.8 x 0.56)
v₂ = 3.31 m/s
Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s
