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?

Question

Marin Koch

Physics

2 September, 16:51

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?

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Elsa Bowers · 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