Question

A block with mass m is attached to a ceiling by a spring with spring constant k and relaxed length . Initially, the spring is compressed to a length of l / 2 . If the block is released, at what distance below the ceiling will the block be brought to rest (instantaneously, at the lowest point) by the spring?

Answer 1

From the conservation of energy:

$$0.5k(0.5l)^2+mgl=-mgx+0.5k(x)^2\\x^2-\frac{2mg}{k}x-\frac{2mg}{k}l-\frac{l^2}{4}=0\\x=\frac{mg}{k}+\sqrt{\frac{m^2g^2}{k^2}+\frac{2mg}{k}l+\frac{l^2}{4}}$$