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A 0.20 kg object, attached to a spring with spring constant k = 10 N/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m. What is its speed at the instant when its displacement is 0.040 m? (Hint: Use conservation of energy.)

Respuesta :

Answer:

 v = 0.566 m / s

Explanation:

Let's use energy conservation at two points the most extreme point of maximum elongation and the point of interest

Initial maximum elongation

    Em₀ = [tex]K_{e}[/tex] = ½ k x²

At this point x = A = 0.080 m

Final point of interest

    [tex]Em_{f}[/tex] = K = ½ m v²

Energy is conserved

   Em₀ = [tex]Em_{f}[/tex]

   ½ k A² = ½ m v²

   v² = k / m A²

   v = √(k/m) A

Let's calculate

   v = √ (10 / 0.20) 0.080

   v = 0.566 m / s

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