Answer to Question #272513 in Physics for Oreoluwa

Question #272513

A 1.50kg object hangs motionless from a spring with a force constant of k=250N/m how far is the spring stretched from its equilibrium length?

Expert's answer

"mg=k\\Delta x\\to \\Delta x=mg\/k=1.5\\cdot9.8\/250=0.0588\\ (m)=5.88\\ (cm)" . Answer

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