A particle of mass m kg is attached to one end A of a model spring OA of natural

Question

A particle of mass m kg is attached to one end A of a model spring OA of natural length I_0= 1 m and stiffness k N/m. The spring is hung vertically with the end O fixed and end A attach to the particle, which moves in a vertical line. The particle's mass, stiffness and natural length of the spring are related to the expression kl_0=2mg: where g is the magnitude of the acceleration due to gravity. It is observed that when the particle is at its highest point during the motion, its distance below O is 1/2 I_0. Using the law of conservation of mechanical energy, find the distance, in m, of the particle below O when it is at its lowest point during the motion. Give your answer to 3 decimal places.

Explanation / Answer

Conserving energy

Decrease in potential energy of mass = increase in PE of spring

mgx = 0.5kx^2

Where x is the distance below A

x = 2mg/k

Given

2mg=kx1

x =1m

Distance from O

D = 1+1 = 2m

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