The length of the unstretched spring is 20 cm, if a weight of 0.5 kg is suspended from it, then the length of the spring will be equal to 0.4 m. Determine the spring stiffness in N / m.
l0 = 20 cm = 0.2 m.
l = 0.4 m.
m = 0.5 kg.
g = 10 m / s2.
When the body is suspended on a spring, then, according to the condition of equilibrium, the action of forces on it is compensated. The body is acted upon by the force of gravity Ft, which is compensated by the force of elasticity of the stretched spring Ff: Ft = Ff.
The force of gravity Ft is determined by the formula: Ft = m * g.
We will express the force of elasticity Fel by Hooke’s law: Fel = k * x.
x = l – l0.
m * g = k * (l – l0).
k = m * g / (l – l0).
k = 0.5 kg * 10 m / s2 / (0.4 m – 0.2 m) = 25 N / m.
Answer: the spring has a stiffness k = 25 N / m.
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