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5 days ago

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An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a dista

nce of 6.4 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?

1 answer:

Sav [1K]5 days ago

6 0

Answer:

h = 12.8 cm

Explanation:

The initial parameters are as follows:

distance = 6.4 cm

when the object descends, its weight matches the spring's force

weight = spring force

mg = ky... equation 1

potential energy stored in a stretched spring = work done by the spring

mgh = 0.5 x k x h^{2}....equation 2

Substituting from equation 1 into equation 2

kyh = 0.5 x k x h^{2}

y = 0.5 x h

2y = h

where y is 6.4, yielding the maximum elongation as

h = 2 x 6.4 = 12.8 cm

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