C. Energy and Momentum · Work, Energy and Simple Harmonic MotionWork Done by a Constant Force in One and Two Dimensions
C2.1 C2.2About 70 minutes
Learning goals- I can calculate the work done by a constant force at any angle to the displacement using W = FΔd cos θ.
- I can decide whether a force does positive, negative or zero work.
- I can find the work done by each force on an object, and the net work, in two-dimensional situations.
- I can connect W = FΔd cos θ to the dot product of force and displacement.
Words to know
work
The energy transferred to an object by a force acting through a displacement: W = FΔd cos θ. Measured in joules (J).
joule (J)
The SI unit of work and energy. 1 J = 1 N·m = 1 kg·m²/s².
net work
The total work done on an object by all the forces acting on it.
component
The part of a vector that lies along a chosen direction, such as F cos θ along the displacement.
scalar
A quantity with size (and possibly sign) but no direction, such as work or energy.
Pulling at an angle
You are pulling a toboggan across a snowy park. If you hold the rope low, almost along the snow, your whole pull drags the toboggan forward. If you lift the rope up high, part of your pull just tries to lift the front of the toboggan. Which way puts more energy into the toboggan for each metre? To answer, you need the full Grade 12 definition of work, which works for a force in any direction.
Remember from Grade 11: in SPH3U (the unit Work, Energy and Power) you used W = FΔd for a force along the line of motion. Work is measured in joules (1 J = 1 N·m). A force along the motion does positive work, a force against the motion does negative work, and a force perpendicular to the motion does no work. This lesson joins those three cases into one formula and uses it in two dimensions.
Work done by a constant force at an angle
When a constant force F acts on an object while the object has a displacement Δd, the work done by that force is
W = FΔd cos θ
- F is the magnitude of the force (N).
- Δd is the magnitude of the displacement (m).
- θ is the angle between the force and the displacement when the two vectors are drawn tail to tail.
Why cos θ? Break the force into two components: F cos θ along the displacement and F sin θ perpendicular to it. Only the component along the motion transfers energy, so W = (F cos θ)Δd.
Work is a scalar. It has a size and a sign but no direction, even though F and Δd are vectors.
The sign of the work comes from cos θ
- 0° ≤ θ < 90°: cos θ is positive, so W is positive. The force adds energy to the object.
- θ = 90°: cos θ = 0, so W = 0. The force changes the direction of motion or holds the object up, but it transfers no energy. The normal force on a level floor and the tension in a string whirling a ball in a horizontal circle are examples.
- 90° < θ ≤ 180°: cos θ is negative, so W is negative. The force removes energy from the object. Kinetic friction (θ = 180°) is the usual example.
In MCV4U (Calculus and Vectors), the unit Dot and Cross Products writes the same idea as a dot product: W = F · Δd. If F = (Fx, Fy) and Δd = (Δx, Δy), then W = FxΔx + FyΔy. Both forms always give the same answer.
Worked exampleExample 1: Toboggan rope at an angle
A parent pulls a toboggan 120 m across a level park. The rope tension is 45 N and the rope makes an angle of 30° above the horizontal. How much work does the rope do? Compare with a horizontal rope.
- The displacement is horizontal and the force is 30° above it, so θ = 30°.
- W = FΔd cos θ = (45 N)(120 m) cos 30°
- W = 5400 × 0.8660 = 4676.5 J ≈ 4.7 × 10³ J.
- With a horizontal rope, θ = 0° and W = (45)(120)(1) = 5400 J = 5.4 × 10³ J.
The angled rope does about 4.7 × 10³ J of work, which is less than the 5.4 × 10³ J of a horizontal rope with the same tension. (Lifting the rope can still help in deep snow, because it reduces the normal force and therefore the friction. Example 2 shows why.)
Worked exampleExample 2: Every force on a sled, in two dimensions
A 15 kg sled is pulled 20 m across level snow by a rope with tension 60 N at 25° above the horizontal. The coefficient of kinetic friction is 0.10. Find the work done by each force and the net work. Use g = 9.8 N/kg.
- Draw a free-body diagram: tension (up and forward at 25°), gravity (down), normal force (up), friction (backward).
- Vertical: the sled does not accelerate up or down, so F_N + F sin 25° = mg. F_N = (15)(9.8) − (60)(sin 25°) = 147 − 25.36 = 121.64 N. (The upward part of the pull lightens the normal force.)
- Friction: F_f = μ_k F_N = 0.10 × 121.64 = 12.164 N.
- Tension: W = (60)(20) cos 25° = +1087.6 J.
- Friction: θ = 180°, so W = (12.164)(20)(−1) = −243.3 J.
- Gravity and normal force: both are perpendicular to Δd, so each does 0 J.
- Net work: 1087.6 − 243.3 = +844.3 J.
Tension +1.09 × 10³ J, friction −2.43 × 10² J, gravity 0, normal 0. Net work +8.44 × 10² J, so the sled speeds up. (The lesson Kinetic Energy and the Work–Energy Theorem in Two Dimensions finds its final speed.)
Worked exampleExample 3: Gravity on a ramp
A 12 kg box slides 3.0 m down a smooth ramp tilted 30° above the horizontal. How much work does gravity do on the box?
- Gravity points straight down. The displacement points down the ramp, 30° below the horizontal.
- The angle between straight down and the ramp direction is 90° − 30° = 60°.
- W = mgΔd cos 60° = (12)(9.8)(3.0)(0.50) = 176.4 J.
- Check another way: the box drops a height Δh = 3.0 sin 30° = 1.5 m, and mgΔh = (12)(9.8)(1.5) = 176.4 J. ✓
Gravity does +1.8 × 10² J of work. It depends only on the vertical drop, an idea you will use in Gravitational Potential Energy and Conservative Forces.
A routine for 2-D work problems
- Sketch the motion and draw Δd as an arrow.
- Draw a free-body diagram. Find any forces you need first (normal force, friction).
- For each force, find the angle θ between it and Δd, tail to tail. Do not just copy an angle from the picture: check which two directions it is between.
- Use W = FΔd cos θ for each force. Keep the signs.
- Add the works for W_net, or use W_net = F_net Δd cos θ, as a check.
Just for funA weightlifter holding 150 kg above her head for ten seconds does zero work on the bar, because it does not move. Physics is strict: you can be exhausted and still have done 0 J.
Hands-on activityMeasure the work of a drag at different angles
You need: a cloth grocery bag holding a few books (about 3 to 5 kg in total), a strong string about 1.5 m long, a luggage scale or spring scale (many homes have a luggage scale), a tape measure, a protractor or a phone protractor app, and a smooth floor with 3 m of clear space.
- Tie the string to the bag handles and hook the scale to the other end.
- Pull the bag slowly at a steady speed for 3.0 m with the string almost horizontal. Read the scale while the bag is moving and record F.
- Repeat with the string at about 30° and then about 50° above the floor (a helper can check the angle with the protractor).
- For each trial calculate W = FΔd cos θ.
- Compare the three readings and the three values of W.
What to look for: the scale reading usually goes down a little at 30°, because lifting part of the weight reduces the normal force and the friction. The work done by the string also usually goes down, because less friction has to be overcome. If you pull very steeply, F rises again. Small differences come from reading a scale on a moving bag; repeat each trial twice and average.
Your turn
1A child pulls a wagon 15 m with a force of 80 N directed 40° above the horizontal. How much work does the child do on the wagon?
W = (80)(15) cos 40° = 1200 × 0.7660 = 919 J ≈ 9.2 × 10² J.
2A 50 N force acts at 90° to an object's 10 m displacement. How much work does it do?
3Kinetic friction of 20 N acts on a box that slides 6.0 m. Find the work done by friction.
Friction is opposite the motion (θ = 180°): W = (20)(6.0)(−1) = −120 J.
4A 100 N force does 1500 J of work on an object that moves 20 m. What angle does the force make with the displacement?
cos θ = W ÷ (FΔd) = 1500 ÷ 2000 = 0.75, so θ ≈ 41° (41.4°).
5A 2.0 kg ball is thrown straight up and rises 5.0 m. How much work does gravity do on it during the rise?
Gravity is down, Δd is up (θ = 180°): W = (2.0)(9.8)(5.0)(−1) = −98 J.
6A satellite moves in a circular orbit around Earth. How much work does gravity do on it during one second? Explain.
0 J. In a circular orbit, gravity points toward the centre and the velocity (and each tiny displacement) is along the circle, so the force is always perpendicular to the motion.
7A 0.50 kg book slides 2.0 m down a smooth 20° incline. How much work does gravity do?
The drop is Δh = 2.0 sin 20° = 0.684 m. W = mgΔh = (0.50)(9.8)(0.684) = 3.4 J.
More practice: warm-up
1What is the SI unit of work, written in base form?
The joule: 1 J = 1 N·m = 1 kg·m²/s².
2F = 10 N and Δd = 3.0 m in the same direction. Find W.
W = (10)(3.0)(cos 0°) = 30 J.
3F = 20 N at 60° to a 5.0 m displacement. Find W.
W = (20)(5.0)(0.5) = 50 J.
4Is work a vector or a scalar?
A scalar. It has a sign but no direction.
5If the angle between F and Δd is 180°, is the work positive, negative or zero?
Negative, because cos 180° = −1.
More practice: on your own
1A shopper pushes a cart 25 m with a force of 40 N directed 30° below the horizontal. How much work does the shopper do?
The force is still 30° from the horizontal displacement: W = (40)(25) cos 30° = 866 J ≈ 8.7 × 10² J. Pushing downward only adds to the normal force.
2A traveller pulls a 25 kg suitcase 12 m across a level floor with a force of 60 N at 35° above the horizontal. μ_k = 0.15. Find the work done by the pull, the work done by friction, and the net work.
F_N = (25)(9.8) − 60 sin 35° = 245 − 34.4 = 210.6 N. F_f = 0.15 × 210.6 = 31.6 N. W_pull = (60)(12) cos 35° = +590 J. W_f = −(31.6)(12) = −379 J. W_net = +211 J.
3You lift a 20 kg box 2.5 m straight up at constant velocity. Find the work done by you, by gravity, and the net work.
Your force equals mg = 196 N, up: W = (196)(2.5) = +490 J. Gravity: −490 J. Net work: 0 J (the speed does not change).
4You push an 18 kg box at constant velocity 3.0 m up a smooth ramp inclined at 25°, pushing parallel to the ramp. How much work do you do?
At constant velocity your push equals mg sin 25° = (18)(9.8)(0.4226) = 74.5 N. W = (74.5)(3.0) = 224 J ≈ 2.2 × 10² J. (Same as mgΔh with Δh = 3.0 sin 25° = 1.27 m.)
5Which does more work over 10 m: a 100 N force at 60° to the motion, or a 60 N force along the motion?
100 N at 60°: (100)(10)(0.5) = 500 J. 60 N along: (60)(10) = 600 J. The 60 N force along the motion does more work.
Challenge
1A force F = (30, 40) N acts on an object while it moves through Δd = (5.0, −2.0) m. Find the work done, and the angle between F and Δd.
W = FxΔx + FyΔy = (30)(5.0) + (40)(−2.0) = 150 − 80 = 70 J. |F| = 50 N, |Δd| = √29 = 5.385 m. cos θ = 70 ÷ (50 × 5.385) = 0.260, so θ ≈ 75°.
2A 30 kg crate is dragged at constant velocity across a level floor by a rope at 30° above the horizontal. μ_k = 0.40. Find the rope tension and the work it does over 10 m.
Horizontal: F cos 30° = μ_k F_N. Vertical: F_N = mg − F sin 30°. So F cos 30° = 0.40(294 − F sin 30°), F(0.866 + 0.200) = 117.6, F ≈ 110 N (110.3 N). W = (110.3)(10) cos 30° = 955 J ≈ 9.6 × 10² J. Friction does −955 J, so the net work is 0, as constant velocity requires.
3A T-bar lift pulls a 70 kg skier 400 m up a slope inclined at 20°, parallel to the slope, at constant speed. Ignore friction. How much work does the lift do? How much work does gravity do?
Pull = mg sin 20° = (70)(9.8)(0.3420) = 234.6 N. W_lift = (234.6)(400) = 9.4 × 10⁴ J. Gravity does −9.4 × 10⁴ J (the skier rises 400 sin 20° = 137 m, and −mgΔh = −9.4 × 10⁴ J).
Quiz
1. A 50 N force acts at 60° to a 4.0 m displacement. How much work does it do?
2. Which force does no work on a car driving along a level road?
3. A box slides across a floor. What is the angle between the kinetic friction force and the displacement?
4. Pulling a sled with a rope angled upward (instead of horizontally) at the same tension changes the friction force because:
5. F = (4, 3) N and Δd = (2, −1) m. What is the work done?
6. A ball on a string moves in a horizontal circle at constant speed. The work done by the string's tension in one revolution is:
Big idea (C2.1, C2.2): work is the energy a force transfers to an object, and only the part of the force along the displacement counts: W = FΔd cos θ. The sign tells whether energy is added (positive) or removed (negative).
How to help: ask your teen to explain, with a sketch, why carrying a box across a room at constant height does no work on the box, but lifting it does. Then pick a household push or pull (a vacuum, a wagon, a laundry basket on a rope) and ask which forces do positive, negative and zero work.
Common difficulties:
- Using the wrong angle. θ must be the angle between F and Δd, tail to tail. On ramps, the angle between gravity and the ramp is 90° minus the ramp angle.
- Forgetting that an angled pull changes the normal force, and so the friction.
- Dropping the negative sign for friction.
- Calculator in radian mode. Physics angles here are in degrees.