Question

A 30 g mass is attached to one end of a 10-cm-long spring.

Answer 1

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;

$\Delta x = L_1 - L_0\\\\\Delta x = 12 \ cm - \ 10 \ cm\\\\\Delta x = 2 \ cm = 0.02 \ m$

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

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

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