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A 0.50-kg block attached to an ideal spring with a spring constant of 80 n/m oscillates on a horizontal frictionless surface. the total mechanical energy is 0.12 j. the greatest extension of the spring from its equilibrium length is:

Respuesta :

jushmk
Work done = 1/2*(max. force - min. force) * greatest extenstion

Max. force = spring constant * greatest extension= 80*d = 80d N
Min. force = spring constant * smallest extenstion = 80*0 = 0 N

Therefore,
Work done = 1/2*80d*d = 40d^2 J

However,
Mechanical energy = Work done
That is,
0.12 = 40d^2
d = Sqrt (0.12/40) = 0.0548 m
The greatest extension from its equilibrium is 0.0548 m.

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