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A 3.00-kg mass hangs vertically downward from a massless spring of spring constant 143.0 N/m. What is the length that the spring stretches from its equilibrium position when the mass is hung from the spring?

Respuesta :

MathPhys

Answer:

0.206 m

Explanation:

According to Hooke's law, the force needed to stretch a spring is equal to the spring constant times the displacement. In this case, the force on the spring is the weight of the mass.

Applying Hooke's law:

F = kx

mg = kx

(3.00 kg) (9.8 m/s²) = (143.0 N/m) x

x = 0.206 m

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