A spring with a spring constant of 110n/m is stretched by 12 cm in the positive direction. How much force is applied to the spring to stretch it?.

Respuesta :

The force applied on the spring to stretch it is 13.2 N.

Hooke's law is a law of elasticity discovered by the English scientist Robert Hooke in 1660 that states that the displacement or size of a deformation is directly proportional to the deforming force or load for relatively small deformations of an object. When the load is removed under these conditions, the object returns to its original shape and size.

According to Hooke's law, F = k*e

where F is the force on the spring

k is force constant

and e is extension

F = (110)*(0.12)

F = 13.2 N

For more information on Hooke's law, visit :

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