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A block of mass m=5.20\;\mathrm{kg}m=5.20kg is being suspended at rest by an extended spring of spring constant k=36.0\;\mathrm{N/m}k=36.0N/m and an external force F=20.0\;\mathrm{N}F=20.0N downward. The spring is attached to the ceiling and has an extended length of s=2.45\;\mathrm{m}s=2.45m. What is the relaxed length s_0s 0 ​ of the spring?

Respuesta :

Answer:

x₀ = 1,894 m

Explanation:

For this exercise let's use Hooke's law

     F = - k Δx = - k ( [tex]x_{f}[/tex] –x₀)

Let's write this equation for the given conditions, the extended length of the spring is the final length (xf = 2.45 m), the sense of force always opposes displacement

With the force applied

         F = -k ( [tex]x_{f}[/tex] - x₀)

        ( [tex]x_{f}[/tex]- x₀) = - F / k

        x₀ =  [tex]x_{f}[/tex] + F / k

        x₀ = 2.45 + (-20.0) / 36.0

        x₀ = 2.45 - 0.556

        x₀ = 1,894 m

This is the so-called natural spring length

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