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A massless spring having a spring constant k = 8.45 N/m is hung vertically. If the spring is displaced 297 mm from its equilibrium position, what is the magnitude of force that the spring exerts?

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The magnitude of the force that a spring exerts when it is displaced can be calculated using Hooke's Law. In this case, with a spring constant of 8.45 N/m and a displacement of 297 mm, the magnitude of the force is 2.511 N.

When a massless spring is displaced from its equilibrium position, it exerts a force that is proportional to the displacement. This relationship is described by Hooke's Law: F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, the spring constant k is given as 8.45 N/m and the displacement x is 297 mm. To find the force, we need to convert the displacement to meters: 297 mm = 297/1000 = 0.297 m. Plugging these values into Hooke's Law, we get F = -8.45 N/m * 0.297 m = -2.511 N.

The magnitude of the force is the absolute value of this result, so the magnitude of the force that the spring exerts is 2.511 N.

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