A 0.16 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 360 N/m.The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest.The block then executes simple harmonic motion along the x-axis (horizontal).When the displacement is x = -0.037 m,the acceleration of the block is closest to: A)85 m/s² B)43 m/s² C)64 m/s² D)270 m/s² E)370 m/s²

Question

Quizplus · Accepted Answer

The acceleration in simple harmonic motion is given by a = -kx/m, where k is the spring constant and m is the mass. Substituting the given values: a = -(360 N/m)(-0.037 m) / 0.16 kg = 83.25 m/s², which is closest to 85 m/s².

A 016 Kg Block on a Horizontal Frictionless Surface Is Attached