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What is the speed of the object?

s is the natural length of the spring, L is the length the spring was extended to. their difference is the extension length x

E=0.5mv^2 and E=0.5kx^2 -> v=√(k(L-s)^2/m)

Now what about vertically? I assume that the speed would be less since there is the gravitational potential that changes as the spring returns to its natural length. Might the answer then be:

E=0.5mv^2 and E=0.5kx^2-mg(L-s) -> v=√[k(L-s)^2-2mg(L-s)]/√m