Question
A .4 kg ball is attached to a spring with spring constant 12 N/m. If the ball is pulled down .2m from the equilibriu, and released, what is its maximum speed while it oscilates?
Answers
(1/2)kX^2 = (1/2) M Vmax^2
Vmx = X*sqrt(k/m)
X is the maximum amount the spring is stretched (or compressed)
Vmx = X*sqrt(k/m)
X is the maximum amount the spring is stretched (or compressed)
Related Questions
"Two springs are hooked together and one end is attached to a ceiling. Spring A has a spring consta...
A spring with spring constant 14.0 N/m hangs from the ceiling. A 490 g ball is attached to the sprin...
A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring
and a...