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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Mộc Miên 4 years 2021-07-19T07:25:50+00:00 1 Answers 23 views 0

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  1. thienhuong
    0
    2021-07-19T07:27:30+00:00
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    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

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