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An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a distance of 3.0 cm below the starting point. If instead of having been lowered slowly the object was dropped from rest, how far then would it then stretch the spring at maximum elongation?

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usmanbiu

Answer:

x = 0.06 m =  6 cm

Explanation:

extension (x) = 3 cm = 0.03 m

force = mg = kx

k (spring constant) = mg / x = mg / 0.03

loss in potential energy = gain in potential energy

mgx = o.5k[tex]x^{2}[/tex]

mg = 0.5kx

x = mg / 0.5k

recall that k = mg / 0.03

therefore

x = [tex]\frac{mg}{\frac{0.5mg}{0.03}}[/tex]

x = mg x [tex]\frac{0.03}{0.5 x mg}[/tex]

x = [tex]\frac{0.03}{0.5}[/tex]

x = 0.06 m =  6 cm

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