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A 30 g mass is attached to one end of a 10-cm-long spring.
The other end of the spring is connected to a frictionless pivot on
a frictionless, horizontal surface. Spinning the mass around in a
circle at 90 rpm causes the spring to stretch to a length of 12 cm.
What is the value of the spring constant?

Sagot :

Onyebuchinnajiidn

The value of the spring constant of the spring is 14.7 N/m.

The given parameters;

  • mass attached to the spring, m = 30 g
  • length of the spring, L₀ = 10 cm
  • angular speed of the mass = 90 rpm
  • final length of the spring, L₁ = 12 cm

The extension of the spring is calculated as follows;

[tex]\Delta x = L_1 - L_0\\\\\Delta x = 12 \ cm - \ 10 \ cm\\\\\Delta x = 2 \ cm = 0.02 \ m[/tex]

The value of the spring constant of the spring is calculated by applying Hooke's law;

[tex]F = mg = k\Delta x\\\\k = \frac{mg}{\Delta x } \\\\k = \frac{0.03 \times 9.8}{0.02} \\\\k = 14.7 \ N/m[/tex]

Learn more about Hooke's law here:https://brainly.com/question/4404276

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