Posted by Tammy on .
An ideal massless spring is fixed to the wall at one end. A block of mass M attached to the other end of the spring oscillates with amplitude A on a frictionless, horizontal surface. The maximum speed of the block is V_m. The force constant of the spring is
a)Mg/A
b)MgV_m/2A
c)MV_m^2/2A
d)MV_m^2/A^2
e)MV_m^2/2A^2
I have narrowed down the choses to either c) d) or e). I don't really get this question.
The max KE of the block is 1/2Mv^2. The max PE of the spring is 1/2 k A^2
set them equal, and solve for k.

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Physics 
Pable,
Because of conservation of energy, kinetic energy K is equal to the potential energy of the spring: K = U. So (mv^2)/2 = (kA^2)/2. Solving for k, we have k = (mv^2)/A^2, which is d).