A 0.20 kg block is attached to a horizontal spring with spring constant 80 N/m. The block is pulled 0.10 m from equilibrium and released from rest on a frictionless surface. What is the speed of the block as it passes through the equilibrium position?
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Correct answer
C. 2.0 m/s
Principle or equation
In the absence of friction, mechanical energy is conserved. The initial elastic potential energy (1/2)kx^2 is converted entirely into kinetic energy (1/2)mv^2 at equilibrium.
Why this answer is correct
Set initial elastic potential energy equal to kinetic energy at equilibrium: (1/2)kx^2 = (1/2)mv^2. Cancel (1/2): kx^2 = mv^2. Solve for v: v = sqrt(kx^2/m) = sqrt(80 × (0.10)^2 / 0.20) = sqrt(0.80/0.20) = sqrt(4.0) = 2.0 m/s.
Example
If a 0.5 kg block is on a spring with k = 200 N/m and compressed 0.05 m, v = sqrt(200×0.0025/0.5) = sqrt(1) = 1 m/s.
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