# *Physics Question* A block of mass m = 0.794 kg is fastened to an unstrained horizontal spring whose spring constant is k = 80.2 N/m.?

A block of mass m = 0.794 kg is fastened to an unstrained horizontal spring whose spring constant is k = 80.2 N/m. The block is given a displacement of +0.130 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. What is the force that the spring exerts on the block just before the block is released?

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- NCSLv 74 years agoFavorite Answer
By Hooke's Law,

F = -kx = -80.2N/m * 0.130m = -10.4 N

The sign indicates "toward the -x axis."

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