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.

Question

3 years ago

8

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?

Physics

1 answer:

kolbaska11 [484] · Answer 1 · 2022-07-29T18:12:17+03:00

kolbaska11 [484]3 years ago

8 0

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