Asked by Mary
A block is suspended from the end of a spring. An external force holds the block so that initially the spring is not stretched or compressed. When the block is released, it oscillates up and down between positions A and B. If the mass of the block is 0.5 kg and the spring constant is 10 N/m, how far will the block fall when it is released?
Answers
Answered by
drwls
It will fall twice the equilibrium stretch distance,
2*M*g/k = 0.98 m, and oscillate about the equilibrium level, M*g/k, eventually settling there.
2*M*g/k = 0.98 m, and oscillate about the equilibrium level, M*g/k, eventually settling there.
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