Figure

Suppose an 0.8 m-long spring has a spring constant of \(k=5\textrm{ N}/\textrm{ m}\).

What spring force is required to compress the spring from its equilibrium position to a length of \(0.5\textrm{ m}\)?
How much work is done by the spring force when compressing the spring from equilibrium to a length of \(0.5\textrm{ m}\)?
How much work is done by the spring force when compressing the spring from 1.2 to \(0.5\textrm{ m}\)?
How much work is done by the spring force when compressing the spring from 1 to \(0.6\textrm{ m}\)?

Solution We begin by positioning an axis parallel to the spring and placing the origin at the free end of the spring in equilibrium. See Figure 57.

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By Hooke's Law, the spring force is \(F=-5x\). When the spring is compressed to a length of \(0.5\textrm{ m}\), then \(x=-0.3\) The spring force \(F\) required to compress the spring to \(0.5\textrm{ m}\) is \[ F=-kx=-5( -0.3) =1.5\hbox{ }\textrm{N} \]
The work \(W\) done by the spring force \(F\) when compressing the spring from equilibrium \((x=0)\) to \(x=-0.3\) is \begin{eqnarray*} W&=&\int_{0}^{-0.3}F(x) {\it dx}\underset{\underset{\color{#0066A7}{\hbox{\(F(x) = -5x\)}}}{\color{#0066A7}{\uparrow }}} {=}\int_{0}^{-0.3}\left( -5x\right) {\it dx} = 5\int_{-0.3}^{0}x\,{\it dx}= 5 \left[\dfrac{x^{2}}{2}\right] _{-0.3}^{0}\\[-11pt] &=&0-\dfrac{5\,\left( -0.3\right) ^{2}}{2}=-0.225\textrm{ J} \end{eqnarray*}
The work \(W\) done by the spring force \(F\) when compressing the spring from \(1.2\textrm{ m}\) \((x=0.4)\) to \(x=-0.3\) is \begin{equation*} W=\int_{0.4}^{-0.3}( -5x) \,{\it dx}=5\int_{-0.3}^{0.4}x\,{\it dx}= 5 \left[\dfrac{x^{2}}{2}\right] _{-0.3}^{0.4}=\dfrac{5}{2}[ 0.4^{2}-( -0.3) ^{2}] =0.175\textrm{ J} \end{equation*}
The work \(W\) done when compressing the spring from \(1\textrm{ m} \ (x=0.2)\) to \(0.6\textrm{ m}\) \((x =-0.2)\) is \[ W=\int_{0.2}^{-0.2}( -5x) \,{\it dx}=5\int_{-0.2}^{0.2}x\,{\it dx}= 5 \left[\dfrac{x^{2}}{2}\right] _{-0.2}^{0.2}=\dfrac{5}{2}[ 0.2^{2}-(-0.2) ^{2}] =0\textrm{ J} \]