Question

A 3500 N is force is applied to a spring that has a spring of constant of k= 14000 N/m. How far from equilibrium will the spring be displaced?

Answer 1

Answer:

the spring be displaced by 25.0 cm

Explanation:

The computation is shown below:

As we know that

F= -K × x

So,

$x = \frac{-F}{K}$

Now  

$x = \frac{-3500}{14000}$

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm