posted by Davis .
A spring with a spring constant k=450.0 N/m is attached to a mass of m=8 kg and stretched -0.5 m from its equilibrium position and released.
What is the speed of the block when it is at the equilibrium position?
When it has stretched from its maximum position, the potential energy stored in the spring is 1/2*k*x^2 = 1/2*450*.5^2 = 56.25
this is converted to kinetic energy:
56.25 = 1/2*m*v^2 = 1/2*8*v^2
Solve for v