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 :
[tex]F = \dfrac{kx^2}{2}[/tex]
Putting all given values, we get :
[tex]F = \dfrac{8.75\times 10^3 \times 0.286^2}{2}\\\\F = 357.86\ N[/tex]
Hence, this is the required solution.
