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A block with mass m =7.4 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.23 m. while at this equilibrium position, the mass is then given an initial push downward at v = 3.9 m/s. the block oscillates on the spring without friction. 1 what is the spring constant of the spring?

Respuesta :

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From the spring equation;

k = -F/x, where k = spring constant, F = weight = mg = 7.4*9.81 = 72.594 N, x = spring displacement = 0.23 m

Substituting;
k = - 72.594/0.23 = -315.63 N/m

The negative sign shows that the force opposes spring constant.

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