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 $W=\int_{0.2}^{-0.2}( -5x) \,{\it dx}=5\int_{-0.2}^{0.2}x\,{\it dx}= 5 \left[\dfrac{x^{2}}{2}\right] _{-0.2}^{0.2}=\dfrac{5}{2}[ 0.2^{2}-(-0.2) ^{2}] =0\textrm{ J}$