D. Energy and Society · Work, Energy and PowerSolving Work Problems: Gravity, Friction, Net Work and Graphs
D2.2 D3.5 D3.2 A1.12About 75 minutes
Learning goals- I can calculate the work done by gravity, by friction and by an applied force on the same object.
- I can find the net work on an object by adding the work done by each force, or from the net force.
- I can find the work done by a constant or changing force from the area under a force–displacement graph.
- I can use free-body diagrams to organize a work problem.
Words to know
net work (W_net)
The total work done on an object by all the forces acting on it. It equals the sum of the work done by each force, and also F_net Δd.
work done against gravity
The work needed to lift an object of mass m through a height Δh at constant velocity: W = mgΔh.
kinetic friction (F_f)
The force that opposes an object sliding over a surface. F_f = μ_k F_N.
force–displacement graph
A graph of force (vertical axis) against displacement (horizontal axis). The area under it equals the work done by the force.
free-body diagram (FBD)
A diagram of one object showing every force acting on it as an arrow from the object.
One crate, four forces
When you drag a crate across a floor, you are not the only thing pushing or pulling on it. Gravity pulls it down, the floor pushes it up and friction drags backward. Each force does its own work: some positive, some negative, some zero. Real work problems ask you to keep track of all of them. A free-body diagram, which you drew in the Forces unit, is the best way to start.
Work done by and against gravity
To lift an object at constant velocity you must push up with a force equal to its weight, F = mg. If it rises a height Δh, the work you do is
W = mgΔh
At the same time gravity pulls down while the object moves up, so gravity does −mgΔh of work. When the object comes down, gravity does +mgΔh.
We use g = 9.8 N/kg (the same number as 9.8 m/s²) near Earth's surface.
Work done by friction
Kinetic friction always acts opposite to the sliding motion, so the work it does is negative:
W_f = −F_f Δd, where F_f = μ_k F_N
On a level surface with no other vertical forces, F_N = mg.
Net work
The net work is the total work done by all the forces on an object. You can find it two ways, and they always agree:
- Add up the work done by each force: W_net = W₁ + W₂ + W₃ + ...
- Use the net force: W_net = F_net Δd (when the net force is along the line of motion)
If the net work is positive, the object speeds up. If it is negative, it slows down. If it is zero (for example, when you push a crate at constant velocity), its speed stays the same. The next lesson, on kinetic energy, shows exactly how much the speed changes.
Worked exampleExample 1: Work done by every force
A 25 kg crate is pulled 8.0 m across a level warehouse floor by a horizontal 120 N force. The coefficient of kinetic friction is 0.30. Find the work done by each force and the net work.
- Draw a free-body diagram: applied force (forward), friction (backward), gravity (down), normal force (up).
- Gravity: F_g = mg = (25 kg)(9.8 N/kg) = 245 N. Level floor, so F_N = 245 N.
- Friction: F_f = μ_k F_N = 0.30 × 245 N = 73.5 N.
- Applied force: W = +(120 N)(8.0 m) = +960 J.
- Friction: W = −(73.5 N)(8.0 m) = −588 J.
- Gravity and normal force: perpendicular to Δd, so 0 J each.
- Net work: 960 − 588 + 0 + 0 = +372 J.
- Check with the net force: F_net = 120 − 73.5 = 46.5 N; W_net = (46.5 N)(8.0 m) = 372 J. ✓
Applied: +960 J; friction: −588 J; gravity and normal: 0 J; net work: +372 J (about +3.7 × 10² J). The crate speeds up.
Work from a force–displacement graph
For a constant force, a graph of F against Δd is a horizontal line, and FΔd is the area of a rectangle under it. This idea works even when the force changes:
The work done by a force equals the area under its force–displacement graph.
Break the area into rectangles and triangles, find each area (in N × m = J) and add. Area below the Δd-axis (a force pointing backward) counts as negative work.
Worked exampleExample 2: Work from a graph
Use the graph above. Find the work done by the force (a) over the first 3.0 m, (b) from 3.0 m to 7.0 m, (c) in total.
- (a) Rectangle: area = base × height = (3.0 m)(40 N) = 120 J.
- (b) Triangle: area = ½ × base × height = ½ (7.0 − 3.0 m)(40 N) = ½ (4.0)(40) = 80 J.
- (c) Total = 120 J + 80 J = 200 J.
120 J, 80 J and 200 J (2.0 × 10² J) in total.
Worked exampleExample 3: Shovelling the driveway
After a snowfall you lift 150 shovelfuls of snow. Each has a mass of 3.0 kg and you lift each one 1.2 m (at constant velocity) to the top of the snowbank. How much work do you do lifting snow? How much work does gravity do on the snow?
- Work for one shovelful: W = mgΔh = (3.0 kg)(9.8 N/kg)(1.2 m) = 35.28 J.
- For 150 shovelfuls: 150 × 35.28 J = 5292 J ≈ 5.3 × 10³ J.
- Gravity acts down while the snow moves up, so gravity does −5.3 × 10³ J.
You do about 5.3 kJ of work lifting the snow; gravity does about −5.3 kJ. (Your body uses much more energy than this, because muscles are far from 100% efficient and you also move the shovel and yourself.)
A problem-solving routine
- Draw a free-body diagram and an arrow for Δd.
- For each force, ask: along Δd (+), opposite Δd (−), or perpendicular (0)?
- Calculate each force you need (mg, μF_N, ...) before calculating work.
- Calculate each work, keep the signs, then add for the net work.
- Check with F_net Δd, and check units (N × m = J).
Just for funShovelling snow is real physics homework: the snow in Example 3 needed over 5000 J of work. That's about the same work as lifting a 540 kg moose up 1 m. (Please do not try to lift the moose.)
Hands-on activityHow much work is your driveway?
You need: a kitchen or bathroom scale, a bucket or pot, a shovel or large scoop, a tape measure, and snow (in warmer months use a pot of water or a bin of sand and lift scoops onto a step).
Grown-up note: shovel only light, fluffy loads; lift with your legs, not your back. Stop if anything hurts.
- Fill the bucket with exactly one shovelful (or scoop) and weigh it on the scale. Subtract the mass of the empty bucket.
- Measure how high you lift each shovelful (from ground to the top of the pile).
- Count how many shovelfuls it takes to clear a section of path (or count how many scoops you lift).
- Calculate the work done on one shovelful and on all of them, using W = mgΔh.
- Sketch a force–displacement graph for lifting one shovelful at constant velocity. What shape is it? What does the area represent?
What to look for: a shovelful of fresh snow is often a few kilograms; wet snow is heavier. A sample: 2.5 kg lifted 0.80 m gives (2.5)(9.8)(0.80) = 19.6 J per load; 200 loads is about 3.9 kJ. The F–d graph is a horizontal line at F = mg, and the rectangle's area is the work.
Your turn
1How much work does gravity do on a 0.50 kg book that falls 1.5 m from a shelf to the floor?
Gravity is along the downward motion: W = mgΔh = (0.50)(9.8)(1.5) = +7.35 J ≈ +7.4 J.
2A 40 kg sled is pulled 30 m across level snow at constant velocity. μ_k = 0.10. Find the work done by friction.
F_N = (40)(9.8) = 392 N; F_f = 0.10 × 392 = 39.2 N. W_f = −(39.2)(30) = −1176 J ≈ −1.2 × 10³ J.
3In the last question, how much work does the pulling force do? What is the net work? Why does that make sense?
At constant velocity the pull equals friction, 39.2 N: W = +1.2 × 10³ J. Net work = 0 J, which makes sense because the speed does not change.
4A constant 15 N force acts over 6.0 m. Describe its F–d graph and find the work from the area.
A horizontal line at F = 15 N from 0 to 6.0 m. Area = rectangle = 15 × 6.0 = 90 J.
5A force rises steadily from 0 N to 50 N over 2.0 m. How much work does it do?
The graph is a triangle: area = ½(2.0 m)(50 N) = 50 J.
6A box has 3 forces doing work on it: +200 J, −80 J and 0 J. What is the net work, and does the box speed up or slow down?
W_net = 200 − 80 + 0 = +120 J. Positive net work, so it speeds up.
More practice: warm-up
1Write the formula for the work needed to lift an object at constant velocity.
2Find the work to lift a 10 kg box 2.0 m at constant velocity.
W = (10)(9.8)(2.0) = 196 J ≈ 2.0 × 10² J.
3What does the area under a force–displacement graph represent?
The work done by the force.
4What is the sign of the work done by kinetic friction?
Negative (friction opposes the motion).
5Name two ways to calculate net work.
Add the work done by each force, or use W_net = F_net Δd.
More practice: on your own
1A 70 kg hiker climbs a trail that rises 350 m in height. How much work does she do against gravity? Does the length of the trail matter?
W = mgΔh = (70)(9.8)(350) = 240 100 J ≈ 2.4 × 10⁵ J. No: the work against gravity depends only on the change in height, not on the path.
2A 60 kg box is pushed 5.0 m across a floor by a 250 N horizontal force. μ_k = 0.35. Find the net work.
F_f = 0.35 × (60)(9.8) = 205.8 N. W_app = 250 × 5.0 = 1250 J; W_f = −205.8 × 5.0 = −1029 J. W_net = +221 J ≈ +2.2 × 10² J.
3A force–displacement graph shows 20 N from 0 to 4.0 m, then −10 N from 4.0 m to 6.0 m. Find the total work.
Area 1 = 20 × 4.0 = 80 J. Area 2 = −10 × 2.0 = −20 J. Total = 60 J.
4A student lowers a 4.0 kg backpack 1.0 m to the floor at constant velocity. Find the work done by gravity and by the student.
Gravity: +mgΔh = +(4.0)(9.8)(1.0) = +39 J. The student pushes up while it moves down: −39 J. Net work = 0 (constant velocity).
5Why do we calculate the friction force before the work done by friction?
Work needs the force in newtons. Friction depends on μ_k and F_N, so you must find F_f = μ_k F_N first, then multiply by Δd.
Challenge
1A 1200 kg car moving along a level road has a net force on it that decreases steadily from 3000 N to 0 N over 50 m. How much net work is done on the car over those 50 m?
The F–d graph is a triangle: W = ½(50 m)(3000 N) = 7.5 × 10⁴ J.
2A 30 kg child slides down a playground slide that is 4.0 m long and drops 2.0 m in height. The friction force along the slide is 60 N. Find (a) the work done by gravity, (b) the work done by friction, (c) the net work.
(a) Gravity's work depends on the height drop: W = mgΔh = (30)(9.8)(2.0) = +588 J. (b) W_f = −(60 N)(4.0 m) = −240 J. (c) W_net = 588 − 240 = +348 J ≈ +3.5 × 10² J (the normal force does no work).
3Explain why the work gravity does on an object depends only on the change in height and not on the path, using a ramp as an example.
Gravity is vertical, so only vertical motion is along its line. On a ramp, the object moves a longer distance, but only the vertical part of that motion counts. Along any path the vertical parts add up to Δh, so W = mgΔh every time. (Formally, the part of gravity along the ramp is smaller by the same factor that the ramp is longer than its height.)
Quiz
1. A 2.0 kg box is lifted 3.0 m at constant velocity. The work done by the lifter is about:
2. The area under a force–displacement graph gives:
3. A crate is pushed at constant velocity across a floor. The net work on it is:
4. A force rises steadily from 0 N to 40 N over 5.0 m. The work done is:
5. A sled slides 10 m. Friction is 25 N. The work done by friction is:
6. Which force does zero work on a box being dragged across a level floor?
Big idea: Real objects have several forces on them, and each does its own work (D2.2, D3.5). The net work tells you whether the object speeds up or slows down, which leads into kinetic energy next lesson. Free-body diagrams (A1.12) keep the work organized.
How to help: Ask your teen to talk you through a free-body diagram for each problem, saying "plus, minus or zero" for each force before any numbers appear.
Common difficulties:
- Multiplying μ by the mass instead of by the normal force in newtons.
- Dropping signs, especially for friction.
- Using the slide or ramp length instead of the height when finding the work done by gravity.
- Graph areas: using base × height for a triangle (forgetting the ½), or not counting area below the axis as negative.
Safety: in the snow activity, keep loads light and lift with the legs. In spring or summer, use water or sand instead.