a mass of 0.5 kg is attached to a spring. the mass is then displace from its equilibrium position by 5cm and released. its speed as it passes the equilibrium position is 50cm/s.

Answer 1

and what is your question?
the max energy in the spring at 5cm is equal to the KE as it passes equilibirum
max energy=1/2 m v^2=1/2 (.5)(.50^2)
so k for the spring must be
PE=1/2 k x^2
k=2(1/2*.5)(.50^2)/(.05^2) in Joules/meter