A 4.8 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface.At time t = 0 s,the block has a displacement of -0.90 m,a velocity of -0.80 m/s,and an acceleration of +2.9 m/s².The force 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

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Quizplus · Accepted Answer

The force constant (k) can be found using Hooke's Law, F = -kx, where F is the force and x is the displacement. The force can also be calculated using F = ma, where m is mass and a is acceleration. Here, F = 4.8 kg * 2.9 m/s² = 13.92 N. Using x = -0.90 m, k = F/x = 13.92 N / 0.90 m ≈ 15.47 N/m, closest to 15 N/m.

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