Question

A force F=0.12N is aplied on spring and spring elongates by 3cm . specific constant of spring ​

Answer 1

The spring constant is 4 N/m

Explanation:

When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:

$F=kx$

where

F is the magnitude of the spring applied

k is the spring constant

x is the elongation of the spring, relative to its equilibrium position

For the spring in this problem, we have:

F = 0.12 N (force applied)

x = 3 cm = 0.03 m (elongation of the spring)

Therefore, we can solve the formula for k to find the spring constant:

$k=\frac{F}{x}=\frac{0.12}{0.03}=4 N/m$

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