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 spring constant $ k $ can be found using the formula $ F = ma = -kx $. Given $ m = 4.8 \, \text{kg} $, $ a = 2.9 \, \text{m/s}^2 $, and $ x = -0.90 \, \text{m} $, we have $ k = \frac{ma}{x} = \frac{4.8 \times 2.9}{0.90} \approx 15 \, \text{N/m} $.

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