13.5: Work done by Non-Constant Forces

Consider a body moving in the x -direction under the influence of a non-constant force in the x -direction, $\overrightarrow{\mathbf{F}}=F_{x} \hat{\mathbf{i}}$. The body moves from an initial position $x_{i}$ to a final position $x_{f}$. In order to calculate the work done by a non-constant force, we will divide up the displacement of the point of application of the force into a large number N of small displacements $\Delta x_{j}$ where the index j marks the $j_{th}$ displacement and takes integer values from 1 to N . Let $\left(F_{x, j}\right)_{\text {ave }}$ denote the average value of the x -component of the force in the displacement interval $\left[x_{j-1}, x_{j}\right]$. For the $j_{th}$ displacement interval we calculate the contribution to the work.

$$W_{j}=\left(F_{x, j}\right)_{\text {ave }} \Delta x_{j} \nonumber$$

This contribution is a scalar so we add up these scalar quantities to get the total work

$$W_{N}=\sum_{j=1}^{j=N} W_{j}=\sum_{j=1}^{j=N}\left(F_{x, j}\right)_{\text {ave }} \Delta x_{j} \nonumber$$

The sum in Equation (13.5.2) depends on the number of divisions N and the width of the intervals $\Delta x_{j}$. In order to define a quantity that is independent of the divisions, we take the limit as $N \rightarrow \infty$ and $\left|\Delta x_{j}\right| \rightarrow 0$ for all j . The work is then

$$W=\lim _{N \rightarrow \infty \atop\left|\Delta x_{j}\right| \rightarrow 0} \sum_{j=1}^{j=N}\left(F_{x, j}\right)_{\text {ave }} \Delta x_{j}=\int_{x=x_{i}}^{x=x_{f}} F_{x}(x) d x \nonumber$$

This last expression is the definite integral of the x -component of the force with respect to the parameter x . In Figure 13.5 we graph the x -component of the force as a function of the parameter x . The work integral is the area under this curve between $x=x_{i}$ and $x=x_{f}$.

Figure 13.5 Plot of x -component of a sample force $F_{x}(x)$ as a function of x .

Example $\PageIndex{1}$: Work done by the Spring Force

Connect one end of an unstretched spring of length $l_{0}$ with spring constant k to an object resting on a smooth frictionless table and fix the other end of the spring to a wall. Choose an origin as shown in the figure. Stretch the spring by an amount $x_{i}$ and release the object. How much work does the spring do on the object when the spring is stretched by an amount $x_{f}$?

Figure 13.6 Equilibrium, initial and final states for a spring

Solution

We first begin by choosing a coordinate system with our origin located at the position of the object when the spring is unstretched (or uncompressed). We choose the $\hat{\mathbf{i}}$ unit vector to point in the direction the object moves when the spring is being stretched. We choose the coordinate function x to denote the position of the object with respect to the origin. We show the coordinate function and free-body force diagram in the figure below.

Figure 13.6a Spring force

The spring force on the object is given by (Figure 13.6a)

$$\overrightarrow{\mathbf{F}}=F_{x} \hat{\mathbf{i}}=-k x \hat{\mathbf{i}} \nonumber$$

In Figure 13.7 we show the graph of the x -component of the spring force, $F_{x}(x)$, as a function of x

Figure 13.7 Plot of spring force $F_{x}(x)$ vs. displacement x

The work done is just the area under the curve for the interval $x_{i}$ to $x_{f}$,

$$W=\int_{x^{\prime}=x_{i}}^{x^{\prime}=x_{f}} F_{x}\left(x^{\prime}\right) d x^{\prime}=\int_{x^{\prime}=x_{i}}^{x^{\prime}=x_{f}}-k x^{\prime} d x^{\prime}=-\frac{1}{2} k\left(x_{f}^{2}-x_{i}^{2}\right) \nonumber$$

This result is independent of the sign of $x_{i}$ and $x_{f}$ because both quantities appear as squares. If the spring is less stretched or compressed in the final state than in the initial state, then the absolute value, $\left|x_{f}\right|<\left|x_{i}\right|$ and the work done by the spring force is positive. The spring force does positive work on the body when the spring goes from a state of “greater tension” to a state of “lesser tension.”