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

the spring be displaced by 25.0 cm

Step-by-step explanation:

The computation is shown below:

As we know that

F= -K × x

So,

`x = (-F)/(K)`

Now

`x = (-3500)/(14000) \\\\`

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm