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## A 0.25 kg mass is placed on a vertically oriented spring that is stretched 0.56 meters from its equilibrium position. If the spring constan

Question

A 0.25 kg mass is placed on a vertically oriented spring that is stretched 0.56 meters from its equilibrium position. If the spring constant is 105 N/m, how fast will the mass be moving when it reaches the equilibrium position? Hint: You cannot ignore the change in gravitational potential energy in this problem. Please give your answer in units of m/s, however, do not explicitly include units when typing your answer into the answer box.

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Physics
7 months
2021-07-19T07:25:50+00:00
2021-07-19T07:25:50+00:00 1 Answers
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## Answers ( )

Answer:The speed of the ball when it reaches equilibrium position is

3.31 m/sExplanation: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/sTherefore, the speed of the ball when it reaches equilibrium position is

3.31 m/s