A 4.8-kg block attached to an ideal spring executes simple harmonic motion on a frictionless horizontal surface.At time t = 0.00 s,the block has a displacement of -0.90m. A velocity of -0.80 m/s And an acceleration of +2.9 m/ s². The force constant (spring constant)of the spring is closest to A)15 N/m. B)14 N/m. C)13 N/m. D)12 N/m. E)11 N/m.

Question

Quizplus · Accepted Answer

The spring constant k can be found using Hooke's law and the formula for acceleration in simple harmonic motion: a = -kx/m. Given a = 2.9 m/s², x = -0.90 m, and m = 4.8 kg, solving for k gives k = (2.9 m/s² * 4.8 kg) / 0.90 m = 15.47 N/m, which is closest to 15 N/m.

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