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A spring has a spring constant of 25 newtons per meter. Calculate the magnitude of the minimum force required to stretch the spring 0.25 meter from its equilibrium position

Answer :

Given data:

* The spring constant of the given spring is k = 25 N/m.

* The stretched length of the spring is x = 0.25 m.

Solution:

The magnitude of minimum force required to stretch the wire is,

[tex]\begin{gathered} F=kx \\ F=25\times0.25 \\ F=6.25\text{ N} \end{gathered}[/tex]

Thus, the magnitude of minimum force required in the given case is 6.25 N.

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