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
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?
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