Work is the mechanical transfer of energy. When you push an object and it moves, you do work on it. When a spring pushes back against your hand, the spring does negative work on you. This section defines work precisely for the simplest case: a force of constant magnitude and direction acting as an object moves.
Geometric definition of work
The work done by a constant force F acting over a displacement Δr depends on three things:
- The magnitude of the force: \( |F| \)
- The magnitude of the displacement: \( |Δr| \)
- The angle θ between the force and the displacement
The work is the dot product of force and displacement:
\[ W = \mathbf{F} \cdot \Delta\mathbf{r} = |\mathbf{F}| |\Delta\mathbf{r}| \cos\theta \]
where F is force [N], Δr is displacement [m], θ is the angle between them [rad], and W is work [J].
Physical meaning: work measures how much of the force acts in the direction of motion. Only the component of force parallel to displacement contributes to work.
\[ F_{\parallel} = F \cos\theta \]
\[ W = F_{\parallel} \cdot \Delta r = (F \cos\theta) \Delta r \]
Sign conventions and energy transfer
The dot product gives work a sign:
-
W > 0(positive work): Force points generally in the direction of motion. The force gives energy to the object. Example: pushing a box across the floor. -
W < 0(negative work): Force points generally against the motion. The force takes away energy from the object. Example: friction opposing a sliding block. -
W = 0(zero work): Force is perpendicular to displacement. No energy transfer. Example: normal force acting on a box moving horizontally.
Important: negative work does not mean "no work" — it means energy is being removed. Friction does negative work on a sliding block, slowing it down.
One-dimensional case (constant force along a line)
In one dimension, if force and displacement both lie on the same line, we can use scalars with signs:
\[ W = F \cdot \Delta x \]
where F and Δx are signed quantities (positive or negative based on direction).
Example 1: A horizontal push
A child pushes a toy cart along a flat street. The push is 50 N horizontal, and the cart moves 20 m in the direction of the push.
\[ W = F \cos(0°) \Delta r = 50 \text{ N} \times \cos(0°) \times 20 \text{ m} = 50 \times 1 \times 20 = 1000 \text{ J} \]
Positive work: the child's push transfers energy to the cart.
Example 2: A push at an angle
A grandfather pushes a lawnmower across a lawn. The push force is 100 N, directed at 30° below the horizontal. The mower moves 30 m horizontally.
\[ W = F \cos(30°) \Delta r = 100 \text{ N} \times \cos(30°) \times 30 \text{ m} \]
\[ = 100 \times 0.866 \times 30 = 2598 \text{ J} \approx 2.6 \text{ kJ} \]
Only the horizontal component (F cos 30°) does work. The downward component increases normal force (and friction), but does no work on the horizontal displacement.
Example 3: Friction opposes motion
A 1500 kg car coasts to a stop on a horizontal road. Friction exerts 2000 N backward (opposite to motion) over a distance of 60 m.
\[ W_{\text{friction}} = F \cos(180°) \Delta r = 2000 \text{ N} \times (-1) \times 60 \text{ m} = -120{,}000 \text{ J} = -120 \text{ kJ} \]
Negative work: friction removes 120 kJ of kinetic energy from the car.
Cumulative work (multiple forces)
When several forces act on an object, the net work is the sum of individual works:
\[ W_{\text{net}} = W_1 + W_2 + W_3 + \cdots \]
Or equivalently, the work done by the net force:
\[ W_{\text{net}} = \mathbf{F}_{\text{net}} \cdot \Delta\mathbf{r} \]
This is crucial for the work–energy theorem (§7.3).
Energy unit: the joule
Work and energy are measured in joules (J):
\[ 1 \text{ J} = 1 \text{ N} \cdot \text{m} = 1 \text{ kg} \cdot \text{m}^2 / \text{s}^2 \]
Real-world scale:
- Climbing one flight of stairs (1.5 m, 70 kg person): ≈ 1 kJ
- Car braking from highway speed to stop: ≈ 1 MJ
- One kilowatt-hour (kWh) = 3.6 MJ (used on electric bills)
Connection to kinematics and dynamics
Recall from Chapter 2 and Chapter 5:
- Displacement \( Δr \) is the change in position over time
- Net force \( F_{\text{net}} \) causes acceleration (Newton's 2nd law: \( F = ma \))
Work bridges kinematics and dynamics: it tells us how much energy is transferred when a force acts through a distance. The next section (§7.2) extends this to variable forces, and §7.3 connects work to the change in kinetic energy.
What you should be able to do
After this section, you should be able to:
- Apply the formula \( W = F \Delta r \cos\theta \) to find work for a constant force at an angle to displacement
- Determine whether work is positive, negative, or zero based on the angle between force and displacement
- Calculate net work when multiple forces act on an object
- Estimate real-world work using the joule scale (kJ, MJ)
- Explain why "only the component of force parallel to displacement matters"
Preview of §7.2
For variable forces — springs, gravity over long distances, position-dependent friction — the force changes as the object moves. We cannot use \( W = F \Delta r \) directly. Instead, we integrate:
\[ W = \int_{\text{path}} \mathbf{F} \cdot d\mathbf{r} \]
This is the focus of §7.2.
📚 See also: Halliday Vol 1, Ch 7, §7.2 — Work Done by a Constant Force. 📖 Open reference: OpenStax University Physics Vol 1 — Chapter 7.1: Work. 📖 Feynman Lectures Vol I — Ch 13: Work and Potential Energy in a Gravitational Field.
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