Suppose a ball with mass M hangs vertically from a spring with stiffness k and relaxed length L0. At what length Leq will the ball encounter equilibrium?

Answer :

Answer:

At equilibrium, the sum of vertical forces are equal to zero. Let's analyze the downward and upward components of the forces.

Downward forces: [tex]F_{down} = Mg[/tex] (Gravity)([tex]g[/tex] is the gravitational constant.)

Upward forces: [tex]F_{up} = kx[/tex] (Spring)

Equilibrium condition: [tex]Mg - kx = 0[/tex].

[tex]x = \frac{Mg}{k}[/tex]

[tex]L_{eq} = L_{0} + x[/tex]

so

[tex]L_{eq} = L_{0} + \frac{Mg}{k}.[/tex]

Explanation:

If there is no masses hanging from the spring, the length of the spring is equal to [tex]L_{0}[/tex]. When there is a mass M hanging, the spring will be stretched by an amount of [tex]x[/tex]. This distance can be found by using the equilibrium condition, that the sum of the forces is equal to zero.

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