Asked by Jay
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?
Answers
Answered by
scott
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
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
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