Question

A block with mass m =7.4 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.22 m. while at this equilibrium position, the mass is then given an initial push downward at v = 3.8 m/s. the block oscillates on the spring without friction. 1 what is the spring constant of the spring?

Answer 1

This is given by F=kx.  Since a force due to gravity is given by F=mg this is F=7.4kg*9.81m/s^2 = 72.594N.  Put this force into the spring equation above and get 72.594N = k*0.22m.  Solve for k to get 329.97N/m