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 $- 0.80 \mathrm {~m} / \mathrm { s } \text {, }$ and an acceleration of $+ 2.9 \mathrm {~m} / \mathrm { s } ^ { 2 }$ The spring constant of the spring is closest to

Question

Quizplus · Accepted Answer

The spring constant $ k $ can be found using Hooke's law $ F = -kx $ and Newton's second law $ F = ma $. Given the acceleration $ a = 2.9 \, \text{m/s}^2 $ and mass $ m = 4.8 \, \text{kg} $, the force is $ F = ma = 4.8 \times 2.9 $. The displacement $ x = -0.90 \, \text{m} $. Solving for $ k $ gives $ k = \frac{ma}{x} = \frac{4.8 \times 2.9}{0.90} \approx 15 \, \text{N/m} $.

A 48-Kg Block Attached to an Ideal Spring Executes Simple Harmonic