An unstretched spring with spring constant 37 N/cm is suspended from the ceiling. A 3.3 kg mass is attached to the spring and let fall. To the nearest tenth of a centimeter, how far does it stretch the spring?

Answer 1

the work stretching the spring equals the change in the gravitational potential energy

1/2 * k * x^2 = m * g * x

37 x^2 = 3.3 * 980 * x = 3234 x

x^2 - (3234/37)x = 0

x = 3234/37