Spring in a motionless elevator

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    Elevator Spring
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To find the spring constant of a spring with a mass of 3.1 kg hanging in a motionless elevator, the forces acting on the mass must be considered. The weight force (w = mg) pulls down, while the spring force (F = -kx) pulls up. Since the elevator is not moving, these forces are equal and opposite, leading to the equation -kx = mg. By substituting the known values, the spring constant can be calculated. This approach clarifies the relationship between the forces and the spring's extension.
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Homework Statement



A single mass m1 = 3.1 kg hangs from a spring in a motionless elevator. The spring is extended x = 11 cm from its unstretched length.
1)What is the spring constant of the spring?

Homework Equations



F=-kx


The Attempt at a Solution



Having trouble setting this one up. I know it is easy but the spring confuses me. I drew a free body diagram and I am still lost.
 
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Hi there! The only forces on the mass will be the weight force (w = mg) pulling down, and the spring force (F = -kx) pulling up. Since the mass is not accelerating, these two forces will be equal and opposite.
 
Yup, got it. -kx=mg. Durrrr. Thanks.
 

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