Question: A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum extension of the spring when the mass is released from rest?
Options:
Correct Answer: 0.5 m
Solution:
Using Hooke\'s law, F = kx, where F = mg = 1 kg * 9.8 m/s² = 9.8 N. Thus, x = F/k = 9.8 N / 200 N/m = 0.049 m.