Question

Calculate the force constant of a spring which is stretched 5 mm by a force of 12 N.
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Answer 1

Answer:

k = 2400 N/m

Explanation:

Given that,

Force acting on the spring, F = 12 N

The spring stretches 5 mm = 0.005 m

We need to find the force constant of the spring.

The force acting on the spring is given by Hooke's law as follows :

F = kx

Where k is force constant

$k=\dfrac{F}{x}\\\\k=\dfrac{12\ N}{0.005\ m}\\\\=2400\ N/m$

So, the force constant of the spring is 2400 N/m.