If the spring constant is 12.6N/m and a spring is stretched by 0.25m, how much force has been applied?

The magnitude of the applied force is 3.15 N

Explanation:

Hooke's law states that for small deformations, the change in dimensions, 'x', of an object is directly proportional to the force, 'F', of the acting load acting that bring about the deformation

Mathematically, Hooke's Law can be expressed as follows;

F = k·x

Where;

F = The applied load

k = The force constant of the object

x = The amount of deformation of the object

The given parameters are;

The spring's force constant, k = 12.6 N/m

The amount of stretching in the spring, x = 0.25 m

Therefore, we have;

The force applied, F = k·x = 12.6 N/m × 0.25 m = 3.15 N.

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