Question

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0905 m from its original length when it reaches equilibrium. The mass is then lifted up a distance L = 0.0225 m from the equilibrium position and released. What is the kinetic energy of the mass at the instant it passes back through the equilibrium position?

Answer 1

at equilibrium, spring force will balance out weight

Kx = Mg ( given data x = 0.0905m , M = 2.15 kg and g = 9.8 m/sec2 )

K = 232.82 N/m

now mass is lifted by 0.0225 m and released, so it will perform SHM ( spring-mass system ) with amplitude = 0.0225m

we know total energy of SHM = 1/2 K A² where A = amplitude  

which remains constant at every point of SHM

and at equilibrium, this energy is completely in form of kinetic energy of the block. ( added gravitational potential energy )

1/2 K A² + mgh = 1/2 m V²

1/2 m v² = (1/2) *232.82 * 0.0225 ² + 2.15 *9.8 *0.0225

V = 0.704 m/sec