A 1kg mass is suspended from a vertical spring with a spring constant of 100 N/m and the equilibrium position is noted. The spring is then pushed upward (compressed) a distance x=10 cm before the mass is released from rest. How fast will the block be moving when it passes through the equilibrium position?

Answer 1

Potential energy stored in the spring during compression will be (1/2)k(0.1)^2
= 0.5*100*10^-2 = 0.5 J. There will be additional stored gravitational energy of 1 kg*9.8 m/s^2*0.1 m = 1.0 J

When the block passes through the equilibrium position, 1.5 will be converted to kinetic energy, (1/2) m V^2

Solve for V