Grade by Grade
C. Geometry and Algebra of Vectors · Dot and Cross Products

Work Done by a Force

C2.4 C2.8About 65 minutes
Learning goals
  • I can explain why the work done by a constant force is the dot product W = F · d.
  • I can calculate work from magnitudes and an angle, or from Cartesian vectors.
  • I can tell when a force does positive, zero or negative work.
  • I can solve work problems involving several forces or an unknown angle.

Words to know

work
The energy transferred when a force acts through a displacement: W = F · d = |F||d| cos θ.
joule (J)
The SI unit of work and energy; 1 J = 1 N·m.
displacement
The change in position, a vector from the start point to the end point.
resultant force
The sum of all forces acting on an object.

Work is a dot product

In Grade 11 Physics (SPH3U, Work, Energy and Power) you learned that a constant force F acting on an object that moves through a displacement d does work W = FΔd cos θ, where θ is the angle between the force and the displacement. Look closely: that is exactly |F||d| cos θ. So

W = F · d

Work is measured in joules (J), where 1 J = 1 N·m. Work is a scalar: it has no direction, which fits perfectly with the dot product giving a number.

Writing work as a dot product has two big advantages. First, it explains the cos θ: only the scalar projection of F on d (the part of the force along the motion) does work. Second, if F and d are given in components, you never need the angle at all.

θF (60 N)long handle: θ = 30°W = 60 × 20 × cos θ ≈ 1039 JdθF (60 N)short handle: θ = 45°W = 60 × 20 × cos θ ≈ 849 Jd
Pulling a wagon 20 m with a 60 N force. Only the horizontal part of the pull (dashed arrow), |F| cos θ, moves the wagon forward. A steeper handle wastes more of the pull lifting the handle, so less work is done per metre.
Worked example

Example 1: Two handle positions

A child pulls a wagon 20 m along a level path with a force of 60 N directed along the handle. Find the work done if the handle makes (a) 30° and (b) 45° with the ground.

  1. W = |F||d| cos θ = (60)(20) cos θ = 1200 cos θ.
  2. (a) 1200 cos 30° = 1200(√3/2) = 600√3 ≈ 1039.2 J.
  3. (b) 1200 cos 45° = 1200(√2/2) = 600√2 ≈ 848.5 J.

(a) about 1039 J, (b) about 849 J. With the same 60 N pull, the lower (longer) handle does more work moving the wagon, because more of the force points along the path. A shorter handle is held at a steeper angle, and more of the pull is spent lifting.

Worked example

Example 2: Cartesian vectors

A force F = [3, 5, −2] N moves an object in a straight line from A(1, 0, 2) to B(6, 4, 3), with distances in metres. Find the work done.

  1. Displacement: d = AB = [6 − 1, 4 − 0, 3 − 2] = [5, 4, 1].
  2. W = F · d = 3(5) + 5(4) + (−2)(1) = 15 + 20 − 2 = 33.

W = 33 J

dFpositive workforce helps the motiondFzero workforce ⊥ motiondFnegative workforce opposes the motion
The sign of W = F · d follows the sign of cos θ. A force along the motion does positive work, a force perpendicular to the motion does no work, and a force against the motion (like friction) does negative work.
Worked example

Example 3: Zero work

You carry a 15 kg backpack 30 m along a level school hallway at a steady height. How much work does gravity do on the backpack?

  1. Gravity points straight down; the displacement is horizontal.
  2. θ = 90°, so W = |F||d| cos 90° = 0.

0 J. (Your arms still get tired, because muscles use energy just to hold a load, but in the physics sense no work is done on the backpack by gravity.)

Worked example

Example 4: Negative work

A hockey puck slides 8 m across the ice, and friction of 0.6 N acts opposite to the motion. Find the work done by friction.

  1. The angle between friction and the displacement is 180°.
  2. W = (0.6)(8) cos 180° = 4.8(−1) = −4.8.

W = −4.8 J. Negative work takes kinetic energy away, which is why the puck slows down.

Worked example

Example 5: Several forces

Two people pull a sled with forces F₁ = [40, 10] N and F₂ = [30, −10] N while it moves through d = [25, 0] m. Find the work done by each force and the total work.

  1. W₁ = [40, 10] · [25, 0] = 1000 J.
  2. W₂ = [30, −10] · [25, 0] = 750 J.
  3. Total: 1000 + 750 = 1750 J.
  4. Check with the resultant force: F₁ + F₂ = [70, 0], and [70, 0] · [25, 0] = 1750 J ✓. This works because of the distributive property: F₁ · d + F₂ · d = (F₁ + F₂) · d.

1000 J, 750 J, total 1750 J.

Worked example

Example 6: Finding the angle

A force of 40 N does 500 J of work moving a crate 15 m. At what angle to the displacement was the force applied?

  1. 500 = (40)(15) cos θ = 600 cos θ.
  2. cos θ = 500/600 ≈ 0.8333.
  3. θ = cos⁻¹(0.8333...) ≈ 33.6°.

θ ≈ 33.6°

Work problems in five steps

  1. Find the displacement vector d (end point minus start point).
  2. If you know components, W = F · d = F₁d₁ + F₂d₂ + F₃d₃.
  3. If you know magnitudes and the angle, W = |F||d| cos θ.
  4. Units: newtons times metres gives joules.
  5. Check the sign: is the force helping or resisting the motion?
Just for fun

Physics fact: if you hold a heavy box perfectly still for an hour, you do zero work on it. Try explaining that to your arms.

Hands-on activity

How much of your push does work?

You need: a protractor (or a free angle-measuring app on a phone), a tape measure and a notebook. Choose three or four household objects that you push or pull with a handle: a vacuum cleaner, a broom, a stroller, a suitcase, a shovel, a wagon or a laundry cart.

  1. Hold each object as you normally do while moving it. Ask a helper to measure the angle θ between the handle and the floor.
  2. For each object, compute cos θ. This is the fraction of your push (along the handle) that acts along the floor and does work moving the object.
  3. Choose one object and move it 5 m. Suppose you push with 30 N along the handle. Compute the work you do: W = (30)(5) cos θ.
  4. Lower or raise the handle by about 15° and repeat. Which way gives more work for the same push?

What to look for: a vacuum cleaner handle is often held at around 45°, so about 71% of the push does work moving it along the floor. With 30 N over 5 m that is about 106 J. Lowering the handle increases cos θ and the work done per newton of push, although pushing down at a steep angle also presses the object into the floor and increases friction.

Your turn

1

A 50 N force acts in the same direction as a 12 m displacement. Find the work.

2

A rope pulls a toboggan 15 m with 80 N at 25° above the horizontal. Find the work.

3

F = [4, 3] N and d = [10, 0] m. Find the work.

4

F = [2, −1, 5] N moves an object from (0, 0, 0) to (3, 4, 1) (metres). Find the work.

5

Friction of 12 N acts against a 5 m displacement. Find the work done by friction.

6

A 25 N force does 300 J of work over 20 m. Find the angle between the force and the displacement.

More practice: warm-up

1

What is the SI unit of work?

2

How much work does a force do if it is perpendicular to the displacement?

3

A 10 N force pushes an object 3 m in the direction of the force. Find the work.

4

Write work as a dot product.

5

Is work a scalar or a vector?

More practice: on your own

1

A toboggan is pulled 40 m across level snow by a rope with tension 70 N at 35° to the ground. Find the work done by the rope.

2

F = [30, −20, 10] N moves an object from A(2, −1, 0) to B(7, 3, 2) (metres). Find the work.

3

A 15 kg box (weight 147 N) is lifted 1.2 m straight up at a steady speed, then carried 10 m horizontally. Find the work done by the lifter's upward force in each part.

4

A lawn mower is pushed 25 m with a force of 120 N along the handle, which points 40° below the horizontal. Find the work done.

5

Forces F₁ = [20, 15] N and F₂ = [−5, 10] N act while an object moves through d = [8, 6] m. Find the total work two ways.

Challenge

1

A wagon must receive 1500 J of work over 25 m, and its handle makes 40° with the ground. What pull along the handle is needed?

2

Prove that a constant force does zero total work on an object that moves around a closed path made of straight segments d₁, d₂, …, dₙ.

3

To keep a sled moving steadily, a 60 N horizontal pull is needed. Compare the pull along a rope, and the work done over 50 m, for a level rope and a rope at 30°.

Check your understanding

Quiz

1.

F = [3, 4] N acts through d = [2, 1] m. How much work is done?

2.

A 100 N force at 60° to the displacement moves an object 5 m. How much work is done?

3.

How much work does a force do when it is perpendicular to the displacement?

4.

Which unit is equivalent to a joule?

5.

A box slides across a floor. What kind of work does the friction force do on the box?

6.

A pull along a wagon handle keeps the same size. Which change increases the work done over the same distance?

For parents and teachers

Big idea (C2.4, C2.8): the work done by a constant force is the dot product W = F · d = |F||d| cos θ, in joules. Only the part of the force along the motion does work. This is the same formula your teen used in Grade 11 Physics, now explained with vectors.

How to help: Do the handle-angle activity together with real household objects. Ask "Does this force do positive, negative or zero work?" for everyday situations: carrying groceries across a parking lot (zero, for the upward force), brakes stopping a car (negative), lifting a bag onto a counter (positive).

Common difficulties:

  • Using the angle with the vertical instead of the angle between the force and the displacement.
  • Forgetting to compute the displacement as end point minus start point.
  • Giving work a direction; it is a scalar.
  • Mixing up "zero work" with "no effort". Holding something still takes effort but does no work on it.