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 $- 0.90 \mathrm {~m} ,$ a velocity of $- 0.80 \mathrm {~m} / \mathrm { s }$ and an acceleration of $+ 2.9 \mathrm {~m} / \mathrm { s } ^ { 2 }$ 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 force constant of the spring is given by:$$k = m\omega^2$$where m is the mass of the block and $\omega$ is the angular frequency of the motion. The angular frequency is given by:$$\omega = \sqrt{\frac{k}{m}}$$Substituting the given values into these equations, we get:$$k = (4.8 \mathrm {~kg}) \left(\sqrt{\frac{2.9 \mathrm {~m} / \mathrm { s } ^ { 2 }}{4.8 \mathrm {~kg}}}\right)^2 = 14.7 \mathrm {~N} / \mathrm { m}$$Therefore, the force constant of the spring is closest to 14 N/m.

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