A particle of mass *m* kg is attached to one
end A of a model spring OA of natural
length *L*_{0} = 0.94 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}=2*m*g, 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 ½ *L*_{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.

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solutionGiven thatA particle of mass i kg is attached to one end 'A' of a model spoing ' OA' of natoral length l_(0)=0.94m and stiffness km//mThe particle's mass, stiffess and natural length of the spring are related to the expression klo=2mgHereg_(1) is the magnitude of the acceroton due to gravityparticle is at highest point during the motion its distance below 0 is 1//2l_(0)Let ' F ' be equilibrium position of ' m ' that{:[k(h-10)=mg],[2k(h-10)=2mg=kl_(0)],[2h-21_(0)=10],[h=(310)/(2)]:}Let ' F ' be convert position of mass im' that is below 'r'matrs{:[([" mechanical "],[" enrigy of ball dtF "],[" enotential "],[" enory in "],[" in "]:}])kinetic enigy of ball of Eo = kinetic ene ... See the full answer