Unit: Work, Energy, and Power

Prerequisites

Later Topics

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Part 1

Vertical mass-spring system: equilibrium

A light vertical spring with force constant kk is fixed to the ceiling. A block of mass mm is attached to the lower end and allowed to hang at rest.

Let the natural (unstretched) length of the spring correspond to y=0y = 0, where yy is the downward extension of the spring from its natural length. Thus the block at rest hangs at y=yeqy = y_{\text{eq}} below y=0y=0.

The situation is shown in the diagram:

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Given: k=80 N/mk = 80\text{ N/m}, m=0.52kgm = 0.52\text{kg}, g=9.8m/s2g = 9.8\text{m/s}^2.

Using force balance in the vertical direction at equilibrium, find the magnitude of the extension yeqy_{\text{eq}} (measured downward from the natural length) first symbolically in terms of mm, gg, and kk, and then numerically.

Correct!

Solution:

Solution

At equilibrium, the acceleration is zero, so in the vertical (yy) direction Newton's second law gives

Fy=may.\sum F_y = m a_y.

At equilibrium ay=0a_y = 0, thus

Fy=0.\sum F_y = 0.

Taking upward as positive yy, the spring force on the block is upward with magnitude Fs=kyeqF_s = k |y_{\text{eq}}|, and the weight is downward with magnitude mgm g.

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Therefore,

Fy=Fsmg=kyeqmg=0.\sum F_y = F_s - m g = k |y_{\text{eq}}| - m g = 0.

Solving for yeq|y_{\text{eq}}| gives

yeq=mgk.|y_{\text{eq}}| = \frac{m g}{k}.

Substituting the given values using the placeholders,

yeq=0.064 m.|y_{\text{eq}}| = 0.064\text{ m}.

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