Physics: what is the change in kinetic energy of the spring?

A 0.51-kg object, suspended from an ideal spring of spring constant 23 N/m, is oscillating vertically. How much change of kinetic energy occurs while the object moves from the equilibrium position to a point 5.4 cm lower? Answer in J



please and Thank youPhysics: what is the change in kinetic energy of the spring?
At the equilibirim position, KE is maxed out. At the apex and trough of the wave, PE is maxed out.



Ignore the Spring Constant



Change in KE = Change in PE = m*g*h = 0.51*9.81*(0.054) = 0.27 JPhysics: what is the change in kinetic energy of the spring?
initial energy = final energy

mgh = 1/2 kx虏 + 1/2 mv虏 where h = x = 0.054m

so

change in KE = mgx - 1/2 kx虏

plug and chug.