Question

A spring has an elastic constant of 8.75 x 103 N/m. What is the magnitude of the force required to stretch this spring a distance of 28.6 cm from its equilibrium position?

Answer 1

Given :

A spring has an elastic constant of 8.75 x 10³ N/m.

To Find :

The magnitude of the force required to stretch this spring a distance of 28.6 cm from its equilibrium position.

Solution :

We know, force required to stretch a spring of spring constant k and distance x is given by :

`F = (kx^2)/(2)`

Putting all given values, we get :

`F = (8.75* 10^3 * 0.286^2)/(2)\\\\F = 357.86\ N`

Hence, this is the required solution.