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A force of 10 N stretches a spring with a spring constant of 20 N/m.

A. Hooke's Law
B. Elastic potential energy
C. Frictional force
D. Inertial motion

1 Answer

4 votes

Final answer:

The answer is A. Hooke's Law.

Step-by-step explanation:

The subject of this question is Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In this case, the force of 10 N is stretching the spring with a spring constant of 20 N/m. This means that the spring will stretch 0.5 meters in response to the applied force.

The equation for Hooke's Law is F = kx, where F is the force, k is the spring constant, and x is the displacement. In this case, we have F = 10 N and k = 20 N/m, so we can rearrange the equation to solve for x: 10 N = 20 N/m * x. Solving for x, we find that x = 0.5 m.

Therefore, the answer to this question is A. Hooke's Law.

The student's question is related to a force of 10 N stretching a spring with a spring constant of 20 N/m. This scenario is described by Hooke's Law, which is F = -kx, where F is the restoring force, k is the force constant, and x is the deformation or displacement of the spring from its equilibrium position.

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