4 bars weighing 100 grams are suspended from a spring, the tension of which is 15 cm.
| June 9, 2021 | education
4 bars weighing 100 grams are suspended from a spring, the tension of which is 15 cm. Determine the tension of the spring when adding a similar block.
m = 100 g = 0.1 kg.
n1 = 4.
n2 = 5.
g = 10 m / s2.
x1 = 15 cm = 0.15 m.
x2 -?
When the bar is in equilibrium, the force of gravity Ft1 is balanced by the force of elasticity of the spring Ffr1: Ft1 = Ffr1.
Fт1 = n1 * m * g.
Fupr1 = k * x1.
n1 * m * g = k * x1.
k = n1 * m * g / x1.
Ft2 = Fcont2.
n2 * m * g = k * x2.
x2 = n2 * m * g / k = n2 * m * g * x1 / n1 * m * g = n2 * x1 / n1.
x2 = 5 * 0.15 m / 4 = 0.1875 m = 18.75 cm.
Answer: the spring will stretch x2 = 18.75 cm.
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