Reading Work From a Graph
A force-position graph can turn a work calculation into a geometric one. Instead of first finding a formula for the force and integrating it, we can use the graph’s shapes: triangles, rectangles, and trapezoids. The key is to interpret area with its sign. A region below the horizontal axis contributes negative work, not positive work.
There is also an important distinction between the graph’s vertical values and the direction in which an object moves. For motion along a position coordinate \(x\), the graph’s vertical values must represent the force component along the positive position coordinate. The displacement direction is accounted for by \(dx\) in the work integral. Thus, a positive graph value does not by itself guarantee positive work if the object moves toward decreasing \(x\).
Signed Area and Work
On a graph with position \(x\) on the horizontal axis and force \(F_x(x)\) on the vertical axis, a region above the horizontal axis has positive signed area. It contributes positive work when the motion traverses that region in the direction of increasing \(x\). A region below the axis has negative signed area and contributes negative work for motion in the direction of increasing \(x\).
For a graph made of straight-line segments, divide the interval at every vertex and every point where the graph crosses the horizontal axis. On each resulting interval, the signed contribution can be calculated from familiar geometric areas. For a linear segment from \((x_1,F_1)\) to \((x_2,F_2)\), its signed trapezoid area is
This formula gives a signed value: \(F_1\) and \(F_2\) retain their signs. If the segment crosses the axis, the formula gives the net contribution on that segment. For a picture-based calculation where the positive and negative regions need to be identified separately, split the segment at its zero crossing.
Why Signed Area Gives Work
The work integral comes from adding the work over many small displacements. Over a short displacement \(\Delta x\), if the force is approximately constant at \(F_x\), the work is approximately \(F_x\Delta x\). Summing these contributions over a partition gives an approximation to the total work. As the partition is made finer, the sum approaches the definite integral.
Each term in the sum is the signed area of a thin rectangle: its height is the force value and its signed width is the displacement. A force value below zero makes the term negative. If the motion is from larger \(x\) to smaller \(x\), the displacement increments are negative, as reflected by reversing the integration bounds. This explains why the graph’s force values should be read along the positive coordinate direction rather than redefined to match the direction of travel.
For a straight-line segment, the graph encloses a trapezoid. Its signed area equals the segment’s contribution to the integral. In fact, if force varies linearly from \(F_1\) to \(F_2\) over a position change \(\Delta x=x_2-x_1\), its average value on that segment is \((F_1+F_2)/2\). Multiplying this average force by \(\Delta x\) gives the trapezoid formula above. This is exact for a linear segment, not an approximation.
Worked Example: A Triangle Followed by a Rectangle
Worked Example: A Triangle Followed by a Rectangle
A force-position graph rises linearly from \(0\text{ N}\) at \(x=0\text{ m}\) to \(4\text{ N}\) at \(x=2\text{ m}\), then stays at \(4\text{ N}\) until \(x=5\text{ m}\). Find the work as the object moves from \(0\text{ m}\) to \(5\text{ m}\).
The first region is a triangle. Its base is \(2\text{ m}\) and its height is \(4\text{ N}\), so its signed area is
The second region is a rectangle of width \(5-2=3\text{ m}\) and height \(4\text{ N}\). Its area is
Both regions lie above the axis, so both contributions are positive. The total work is
Check by integration. On \(0\le x\le2\), the line has slope \(4/2=2\text{ N/m}\), so \(F_x(x)=2x\). On \(2\le x\le5\), \(F_x(x)=4\). Therefore
The integral agrees with the geometric calculation, including the units \(\text{N}\cdot\text{m}=\text{J}\).
When the Graph Crosses the Axis
A force can change sign during a displacement. The point where the graph crosses the horizontal axis is where the force component is zero. Locate that point before calculating the regions. Areas above and below the axis must then be added with opposite signs.
Worked Example: A Linear Force That Changes Sign
Between \(x=0\text{ m}\) and \(x=2\text{ m}\), the force decreases linearly from \(3\text{ N}\) to \(-1\text{ N}\). Find the work for motion from \(0\text{ m}\) to \(2\text{ m}\).
The slope is
Since the force is \(3\text{ N}\) at \(x=0\), its equation is \(F_x(x)=3-2x\). The zero crossing satisfies \(3-2x=0\), so \(x=1.5\text{ m}\).
From \(0\) to \(1.5\text{ m}\), the graph forms a positive triangle with height \(3\text{ N}\):
From \(1.5\) to \(2\text{ m}\), the graph forms a negative triangle with depth \(1\text{ N}\):
Thus the net work is
Check directly with the integral:
The result is positive because the positive triangular contribution is larger than the negative one. The check confirms the signed-area calculation.
Piecewise-Linear Graphs: A Reliable Procedure
For a graph made of connected straight-line segments, the main challenge is organizing the regions without losing a sign or interval. A systematic procedure is:
Record the starting and ending positions. The integral runs from the starting position to the ending position.
These points divide the graph into simple regions and identify where the sign of force changes.
Use triangle, rectangle, or trapezoid formulas, assigning a negative sign below the axis.
Compare with the integral of the corresponding line equations when available, and verify that the units are joules.
Alternatively, on a linear segment with endpoint forces \(F_i\) and \(F_{i+1}\), use the signed trapezoid contribution \(\frac{F_i+F_{i+1}}{2}(x_{i+1}-x_i)\). If the segment crosses the axis, splitting it into two triangles makes the separate positive and negative contributions explicit.
Worked Example: Several Segments and Two Sign Changes
A piecewise-linear graph passes through \((0,-2)\), \((2,2)\), \((4,2)\), and \((6,-2)\), where position is in metres and force is in newtons. Find the work from \(x=0\text{ m}\) to \(x=6\text{ m}\).
On the first segment, the force rises from \(-2\text{ N}\) to \(2\text{ N}\) over \(2\text{ m}\). It crosses zero halfway, at \(x=1\text{ m}\). The negative triangle contributes
and the positive triangle from \(1\) to \(2\text{ m}\) contributes
The net contribution of the first segment is \(-1+1=0\text{ J}\).
The middle segment is a rectangle at \(2\text{ N}\) from \(x=2\) to \(x=4\). Its work is
The final segment falls from \(2\text{ N}\) to \(-2\text{ N}\) over \(2\text{ m}\), crossing zero at \(x=5\text{ m}\). Its positive triangle contributes \(1\text{ J}\), and its negative triangle contributes \(-1\text{ J}\), so this segment contributes \(0\text{ J}\).
Adding all signed contributions gives
Check using piecewise equations. The first line is \(F_x(x)=2x-2\) on \(0\le x\le2\), the middle force is \(2\) on \(2\le x\le4\), and the final line is \(10-2x\) on \(4\le x\le6\). Hence
This agrees with the graph. The first and last segments each have equal positive and negative triangular areas, while the middle rectangle supplies all the net work.
Motion Toward Decreasing Position
The signed area rule must be applied together with the direction of the integration limits. If \(F_x\) is positive but the object moves from a larger position to a smaller one, then \(dx\) is negative overall and the work is negative. Do not change the sign of the graph’s force values just because the object moves left.
Worked Example: Positive Force During Decreasing Position
A constant force component \(F_x=5\text{ N}\) acts while an object moves from \(x=1\text{ m}\) to \(x=0\text{ m}\). Calculate the work.
The graph has height \(5\text{ N}\), and the directed change in position is \(0-1=-1\text{ m}\). Therefore
Equivalently, reversing the bounds changes the sign:
The force graph remains above the axis because the force component along positive \(x\) is \(+5\text{ N}\). The negative work comes from the motion toward decreasing \(x\), not from changing the graph’s force value. Both calculations give \(-5\text{ J}\).
Common Pitfalls
A frequent error is to add the ordinary, nonnegative areas of every region and call the result work. For example, in the sign-changing example, the two geometric areas are \(2.25\text{ J}\) and \(0.25\text{ J}\). Adding them gives \(2.50\text{ J}\), but that is not the net work. The negative region must be subtracted:
Another error is to assume that a positive force graph always means positive work. That is true for motion over the same interval in the increasing-\(x\) direction, but not when the limits are reversed. Keep the force component tied to the positive coordinate, preserve the direction in the limits, and assign signs to areas consistently.
Finally, check dimensions. A graph’s area has units of force times position, such as \(\text{N}\cdot\text{m}\), which equals joules. If the calculation gives only newtons or only metres, a width, height, or unit conversion has been omitted.
Check Your Understanding
Use signed areas and directed bounds in each question.
- A force rises linearly from \(0\text{ N}\) at \(x=0\text{ m}\) to \(6\text{ N}\) at \(x=3\text{ m}\). What work is done from \(0\) to \(3\text{ m}\)?
- A graph has a positive triangular area of \(8\text{ J}\) and a negative triangular area of magnitude \(3\text{ J}\). What is the net work?
- A constant force component of \(-4\text{ N}\) acts as position increases from \(x=1\text{ m}\) to \(x=5\text{ m}\). What is the work?
- Why must a straight-line segment that crosses the horizontal axis be split when identifying its positive and negative contributions?
- A constant force component of \(7\text{ N}\) acts as an object moves from \(x=3\text{ m}\) to \(x=1\text{ m}\). What is the work, and why is it negative?
