Answered

Get the information you need quickly and easily with IDNLearn.com. Our community is here to provide detailed and trustworthy answers to any questions you may have.

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

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for choosing IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more solutions.