contestada

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?

Respuesta :

Answer:

the spring be displaced by 25.0 cm

Explanation:

The computation is shown below:

As we know that

F= -K × x

So,

[tex]x = \frac{-F}{K}[/tex]

Now  

[tex]x = \frac{-3500}{14000} \\\\[/tex]

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm

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