C. Energy and Momentum · Work, Energy and Simple Harmonic MotionWork Done by a Varying Force: The Area Under the Graph
C2.2 C2.1 A1.8About 65 minutes
Learning goals- I can find the work done by a force as the area under its force–displacement graph.
- I can treat area below the axis as negative work.
- I can estimate work from a curved graph or a data table using trapezoids or counting squares.
- I can explain why the area method works for a changing force.
Words to know
force–displacement graph
A graph of the force component along the motion (vertical axis) against position or displacement (horizontal axis).
area under a graph
The area between a graph and the horizontal axis over an interval; for an F–Δd graph it equals the work.
trapezoid rule
An estimate of an area made by joining neighbouring data points with straight lines and adding the trapezoid areas ½(F₁ + F₂)Δx.
variable force
A force whose size or direction changes as the object moves, such as the force of a stretched cord.
When the force keeps changing
Stretch a bungee cord, draw a bow, or squeeze a tennis ball. The harder you pull or press, the harder it pushes back. The force is not constant, so W = FΔd cos θ cannot be used in one step: which F would you put in? This lesson shows the tool physicists use for any force, constant or not: the area under a force–displacement graph.
Remember from Grade 11: in SPH3U you found work as the area under simple force–displacement graphs made of rectangles and triangles. Here you will extend that to curved graphs, data tables, negative areas and estimates.
Work is the area under the F–Δd graph
Plot the component of the force along the motion (F cos θ, or just F when the force is along the motion) on the vertical axis, and the position or displacement on the horizontal axis.
The work done by the force between two positions equals the area between the graph and the Δd-axis over that interval.
Why? Over a very small displacement the force is almost constant, so the work there is a thin rectangle: F × (small Δd). Adding all the thin rectangles gives the whole area.
- Area above the axis (force along the motion) is positive work.
- Area below the axis (force against the motion) is negative work.
- The units of area are N × m = J, so read the axis units carefully (kN, cm and mm are common traps).
Worked exampleExample 1: Work from a piecewise graph
Use the graph above. The force acts along the motion. Find the work done (a) from 0 m to 2 m, (b) from 2 m to 5 m, (c) from 0 m to 7 m.
- (a) Triangle: ½ × base × height = ½ (2.0 m)(30 N) = 30 J.
- (b) Rectangle: (5.0 − 2.0 m)(30 N) = 90 J.
- (c) The last piece, 5 m to 7 m, is another triangle: ½ (2.0)(30) = 30 J. Total = 30 + 90 + 30 = 150 J.
(a) 30 J, (b) 90 J, (c) 1.5 × 10² J.
Worked exampleExample 2: A force that changes direction
A robot vacuum is pushed by a force of +20 N for the first 4.0 m of its motion. Then a stronger obstacle pushes back on it with −10 N over the next 2.0 m (it keeps moving forward). Find the net work done by these forces.
- 0 to 4.0 m: rectangle above the axis, (4.0)(20) = +80 J.
- 4.0 to 6.0 m: rectangle below the axis, (2.0)(−10) = −20 J.
- Total: 80 + (−20) = +60 J.
+60 J. Area below the axis subtracts.
Worked exampleExample 3: A force that grows steadily
The force needed to stretch a cord grows in proportion to the stretch: F = (200 N/m)x. How much work does it take to stretch the cord (a) from 0 to 0.30 m, (b) from 0.30 m to 0.50 m?
- The graph is a straight line through the origin. At x = 0.30 m, F = 60 N. At x = 0.50 m, F = 100 N.
- (a) Triangle: ½ (0.30 m)(60 N) = 9.0 J.
- (b) Trapezoid: ½ (60 N + 100 N)(0.20 m) = 16 J.
- Notice the second stretch is shorter (0.20 m) but needs more work, because the force is larger there.
(a) 9.0 J, (b) 16 J. This straight-line force is exactly the kind of force a spring exerts. In Hooke's Law: Springs and Restoring Forces you will turn this area into a formula.
Curved graphs and data tables
Real force graphs are often curved, or you only have measurements at a few positions. Then you estimate the area:
- Count squares. Find the work represented by one grid square (for example 0.1 m × 10 N = 1 J), count full squares and add up partial ones.
- Trapezoids. Join neighbouring data points with straight lines and add the trapezoid areas: ½(F₁ + F₂)Δx for each strip. Narrower strips give better estimates.
In first-year university physics, and in the integral calculus that follows MCV4U, the exact area is written as a definite integral, W = ∫F dx. The idea is the same: add up F × (tiny Δx) over the whole motion.
Worked exampleExample 4: Estimating from data
A student measures the force needed to draw a practice bow (data in the graph above): at draws of 0, 0.1, 0.2, 0.3, 0.4 and 0.5 m the force is 0, 60, 110, 150, 180 and 200 N. Estimate the work done drawing the bow 0.50 m.
- Use trapezoids 0.1 m wide: ½(F₁ + F₂)(0.1 m) for each strip.
- Strips: ½(0 + 60)(0.1) = 3 J; ½(60 + 110)(0.1) = 8.5 J; ½(110 + 150)(0.1) = 13 J; ½(150 + 180)(0.1) = 16.5 J; ½(180 + 200)(0.1) = 19 J.
- Sum: 3 + 8.5 + 13 + 16.5 + 19 = 60 J.
About 60 J. The graph curves slightly downward (it is concave down), so the straight tops of the trapezoids lie a little below the curve and 60 J is a slight underestimate.
Just for funWhy did the archer spend the weekend studying force–displacement graphs?
She wanted to know how much work goes into every draw.
Hands-on activityMake your own force–stretch graph
You need: a long, wide rubber band (or a thin elastic exercise band), a plastic grocery bag, 4 or 5 identical cans of food, a kitchen scale, a ruler, tape, and a sturdy place to hang the band, such as a cupboard door handle. Safety: keep your face away from the stretched band in case it snaps, and do not overload it.
- Weigh one can on the kitchen scale (for example 0.40 kg). Its weight is F = mg (0.40 kg × 9.8 N/kg = 3.9 N).
- Hang the bag from the band. Tape the ruler beside it and record the position of the bottom of the band with the empty bag.
- Add one can at a time. Each time record the total force (number of cans × weight of one can) and the stretch x from the starting position.
- Plot F (vertical) against x (horizontal) on graph paper.
- Estimate the work done stretching the band to its largest stretch, using triangles or trapezoids.
What to look for: a rubber band's graph is often close to a straight line for small stretches and then curves. Your work should come out to a few tenths of a joule up to about 1 J, depending on the band. Check your units: stretch in metres, force in newtons.
Your turn
1A constant 25 N force acts along a 6.0 m displacement. Describe its F–Δd graph and find the work as an area.
A horizontal line at 25 N from 0 to 6.0 m. Area of the rectangle: (25)(6.0) = 150 J.
2A force rises steadily from 0 to 40 N over 0.50 m. Find the work.
Triangle: ½(0.50)(40) = 10 J.
3A force is 12 N from 0 to 3.0 m, then drops instantly to 4.0 N from 3.0 m to 5.0 m. Find the total work.
(12)(3.0) + (4.0)(2.0) = 36 + 8 = 44 J.
4A force decreases steadily from 30 N at 0 m to 10 N at 4.0 m. Find the work.
Trapezoid: ½(30 + 10)(4.0) = 80 J.
5On an F–Δd graph, a region of area 15 J lies below the axis and a region of area 40 J lies above it. What is the total work?
6A graph's horizontal axis is in centimetres and its vertical axis in newtons. A triangle under it has base 20 cm and height 50 N. Find the work in joules.
Convert: 20 cm = 0.20 m. ½(0.20)(50) = 5.0 J.
7Why is the area under a force–displacement graph equal to work, even when the force changes?
Over a tiny displacement the force is nearly constant, so each thin strip has area F × Δd, which is the work over that tiny step. Adding all the strips adds all the small works, giving the total work.
More practice: warm-up
1What quantity does the area under a force–displacement graph represent?
The work done by the force.
2What units does the area have if F is in N and Δd is in m?
3Is area below the Δd-axis positive or negative work?
4A rectangle under an F–Δd graph is 2.0 m wide and 15 N tall. Find the work.
5A triangle under an F–Δd graph has base 4.0 m and height 10 N. Find the work.
More practice: on your own
1A force grows from 0 to 50 N over the first 1.0 m, stays at 50 N until 3.0 m, then falls to 0 at 4.0 m. Find the total work.
Triangle ½(1.0)(50) = 25 J; rectangle (2.0)(50) = 100 J; triangle ½(1.0)(50) = 25 J. Total 150 J.
2A spring-loaded toy needs F = (500 N/m)x to compress it. Find the work to compress it 4.0 cm.
x = 0.040 m, F = 20 N. Triangle ½(0.040)(20) = 0.40 J.
3The same toy is compressed from 4.0 cm to 8.0 cm. Find the extra work, and compare it with your last answer.
At 8.0 cm, F = 40 N. Trapezoid ½(20 + 40)(0.040) = 1.2 J, three times the work for the first 4.0 cm, even though the distance is the same.
4A force–displacement table reads (0 m, 0 N), (1 m, 8 N), (2 m, 12 N), (3 m, 14 N). Estimate the work from 0 to 3 m with trapezoids.
½(0 + 8)(1) + ½(8 + 12)(1) + ½(12 + 14)(1) = 4 + 10 + 13 = 27 J.
5A puck is pushed by +6.0 N for 0.50 m and then rubbed backward by −2.0 N for 1.5 m while still moving forward. Find the net work.
(6.0)(0.50) + (−2.0)(1.5) = 3.0 − 3.0 = 0 J. The puck ends with the same speed it started with.
Challenge
1A force acts at 37° to the displacement. Its magnitude rises steadily from 0 to 50 N over 6.0 m. Find the work.
Only the component along the motion counts: it rises from 0 to 50 cos 37° = 39.9 N. Area: ½(6.0)(39.9) = 120 J (1.2 × 10² J).
2The force on a drawn bow is modelled by F = 500x − 400x² (F in N, x in m) for 0 ≤ x ≤ 0.50 m. Estimate the work from 0 to 0.50 m with five trapezoids 0.1 m wide. The exact value (by integration) is 45.83 J. Is your estimate too high or too low, and why?
F at 0, 0.1, 0.2, 0.3, 0.4, 0.5 m: 0, 46, 84, 114, 136, 150 N. Trapezoids: 0.1[(0 + 150)/2 + 46 + 84 + 114 + 136] = 0.1(75 + 380) = 45.5 J. This is slightly too low: the graph is concave down, so the straight trapezoid tops lie below the curve.
3A 2.0 kg cart starts at rest. A horizontal force on it is 10 N for the first 1.0 m and then decreases steadily to 0 at 3.0 m. Ignoring friction, how much work does the force do, and what is the cart's speed at 3.0 m? (Use Ek = ½mv² from Grade 11.)
W = (10)(1.0) + ½(2.0)(10) = 10 + 10 = 20 J. All of it becomes kinetic energy: 20 = ½(2.0)v², v² = 20, v ≈ 4.5 m/s.
Quiz
1. The area under a force–displacement graph represents:
2. A force rises steadily from 0 to 80 N over 0.25 m. How much work does it do?
3. On an F–Δd graph, part of the area lies below the Δd-axis. That part means:
4. A graph uses kN for force and cm for displacement. A rectangle under it is 2.0 kN tall and 30 cm wide. What is the work?
5. Measurements of a changing force are taken every 0.1 m. Which method gives a good estimate of the work?
6. A force F = (300 N/m)x stretches a cord from x = 0 to x = 0.20 m. The work done is:
Big idea (C2.2): for any force, even one that changes as the object moves, the work done equals the area under the force–displacement graph. Students break the area into rectangles, triangles and trapezoids, or estimate it from data.
How to help: do the rubber-band activity together. Ask your teen to show you which shape on their graph represents the work, and to explain why the first centimetre of stretch takes less work than the last centimetre.
Common difficulties:
- Forgetting the ½ in triangle and trapezoid areas.
- Not converting cm to m, or kN to N, before multiplying.
- Treating area below the axis as positive.
- Plotting the full force instead of its component along the motion when the force is at an angle.