A 0.28-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 500 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = -0.052 m, find the acceleration of the block. A) 92 m/s² B) 46 m/s² C) 69 m/s² D) 230 m/s² E) 280 m/s²

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Quizplus · Accepted Answer

The acceleration in simple harmonic motion is given by a = -kx/m. Here, k = 500 N/m, x = -0.052 m, and m = 0.28 kg. Substituting these values gives a = -(500)(-0.052)/0.28 = 92.857 m/s², which rounds to 92 m/s².

A 028-Kg Block on a Horizontal Frictionless Surface Is Attached to to an Ideal