A block of mass m = 0.775 kg is fastened to an unstrained horizontal spring whose spring constant is k = 83.6 N/m. The block is given a displacement of + 0.113 m, where the + sign indicates that the displacement is along the + x axis, and then released from rest. What is the force (magnitude) that the spring exerts on the block just before the block is released?

Question

Physics

Izabelle Kaiser

12 October, 05:47

A block of mass m = 0.775 kg is fastened to an unstrained horizontal spring whose spring constant is k = 83.6 N/m. The block is given a displacement of + 0.113 m, where the + sign indicates that the displacement is along the + x axis, and then released from rest. What is the force (magnitude) that the spring exerts on the block just before the block is released?

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Answers (1)

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Rodney Sherman · Answer 1

F = 9.45 N

Explanation:

If the mass is fastened to an unstrained horizontal spring, this means that at this position, the spring doesn't exert any force, because it keeps his equilibrium length.

If then the block is given a displacement of + 0.113m, this means that the spring has been stretched in the same length.

According to Hooke's Law, the spring exerts a restoring force (trying to return to his equilibrium state) that opposes to the displacement, and which is proportional (in magnitude) to it, being the proportionality constant, a quantity called spring constant, which depends on the type of spring.

We can write the Hooke's Law as follows:

F = - k * Δx

Just before the block is released, we can get the value of F as follows:

⇒ F = 83.6 N/m * 0.113 m = 9.45 N (in magnitude)