Answer to Question #272513 in Physics for Oreoluwa

Question

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?

Answer 1

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