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HaoPhysics
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Homework Statement
5 identical masses M are suspended by a spring stretched a distance of L. If 3 of these masses are removed, what is the potential energy stored in the spring?
2. Relevant diagram
So L is the distance stretched with 5 masses.
Let L2 be the distance stretched with 2 masses remaining.
L2 < L since the spring would "shrink" if it had to hold up less mass.
L2 = (2/5)L
The potential energy is E = ½(k*x2)
The Attempt at a Solution
The answer which the book gives is (5/2)MgL
Which would make sense if we are measuring the potential energy when the spring is holding 5 masses.
Since F = 5Mg = kx, and x = L, so E = ½(k*x2) = ½(5MgL) = (5/2)MgL
But isn't the question asking the potential energy when the spring is holding 2 masses?
As I understand it, the potential energy would change:
F = 2Mg = kx, and x = (2/5)L, so E = ½(k*x2) = ½(4/5MgL)=(2/5)MgL
Which answer should it be?
Which answer should it be?