Question
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
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
Related Questions
One end of a massless coil spring is suspended from a rigid ceiling. It is found that when a 4 kg...
A spring is hung from a ceiling, and an object attached to its lower end stretches the spring by a d...
An ideal spring has a spring constant k = 20 N/m. The spring is suspended vertically. A 0.7-kg body...
A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring
and a...