A spring is hung from the ceiling. When a 0.450 kg block is attached to the free end of the spring and released from rest, the block drops 5.0 cm before momentarily coming to rest, after which it moves back upward. What is the spring constant of the spring?

answer

88.2  N/m

set up equation

when the block drops to 5.0 cm, the weight of block and the spring force on the block are equal because there is no acceleration on the block (at rest)

so mg = Fs

due to Hooke's law, Fs = kx

where k is the spring constant and x is the displacement

mg = Fs = kx

mg = kx

values

convert given values to standard units

m = 0.450 kg

g = 9.8 m/s^2

x = 5.0 cm = 0.05 m

plug in values and solve

mg = kx

0.45 * 9.8 = 0.05x

x = 0.45 * 9.8 / 0.05

x = 88.2  N/m

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you can cut your hand off if your not