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ki77a [65]
3 years ago
12

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

ll encounter equilibrium?
Physics
1 answer:
Eddi Din [679]3 years ago
6 0

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: F_{down} = Mg (Gravity)(g is the gravitational constant.)

Upward forces: F_{up} = kx (Spring)

Equilibrium condition: Mg - kx = 0.

x = \frac{Mg}{k}

L_{eq} = L_{0} + x

so

L_{eq} = L_{0} + \frac{Mg}{k}.

Explanation:

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

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