A 0.80 kg object is attached to a horizontal spring with spring constant 200 N/m. The object is pulled 0.15 m from equilibrium and released from rest on a frictionless surface. What is the speed of the object when it passes through the equilibrium point?
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Correct answer
B. 2.4 m/s
Principle or equation
Conservation of mechanical energy: the elastic potential energy at maximum displacement (1/2 k x²) is converted entirely into kinetic energy (1/2 m v²) at equilibrium.
Why this answer is correct
Set 1/2 k x² = 1/2 m v². Solve for v: v = x sqrt(k/m) = 0.15 × sqrt(200/0.80) = 0.15 × sqrt(250) ≈ 0.15 × 15.81 ≈ 2.37 m/s, which rounds to 2.4 m/s.
Example
If k = 100 N/m, m = 0.5 kg, x = 0.1 m, then v = 0.1 × sqrt(100/0.5) = 0.1 × sqrt(200) ≈ 1.41 m/s.
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