Introduction
Work in physics has a precise meaning different from everyday usage. Work is done when a force causes displacement in its direction. This scalar quantity is fundamental to understanding energy transfer.
Before studying this topic, make sure you understand:
- Vectors and Dot Product - Work uses the scalar product
- Newton’s Laws of Motion - Understanding forces
Work by a Constant Force
Definition
Work done by a constant force $\vec{F}$ over displacement $\vec{s}$:
$$\boxed{W = \vec{F} \cdot \vec{s} = Fs\cos\theta}$$where $\theta$ is the angle between $\vec{F}$ and $\vec{s}$.
Understanding the Formula
- $F$ = magnitude of force (N)
- $s$ = magnitude of displacement (m)
- $\cos\theta$ = determines how much of the force is in the direction of motion
- Unit: Joule (J) = N·m
Work equals the component of force along the displacement times the displacement:
$$W = (F\cos\theta) \times s = F_{\parallel} \times s$$Interactive Demo: Visualize Work Done
Explore how force and displacement combine to produce work.
Sign Convention
The sign of work tells us about energy transfer:
| Condition | $\cos\theta$ | Work | Meaning |
|---|---|---|---|
| $\theta < 90°$ | Positive | Positive | Force aids motion, energy added |
| $\theta = 90°$ | Zero | Zero | Force perpendicular to motion |
| $\theta > 90°$ | Negative | Negative | Force opposes motion, energy removed |
Examples
- Lifting an object: Work by you is positive, work by gravity is negative
- Pushing a box on floor: Work by push is positive, work by friction is negative
- Circular motion: Work by centripetal force is zero (always perpendicular)
Work by Variable Force
When force varies with position, we use integration:
$$\boxed{W = \int_A^B \vec{F} \cdot d\vec{r}}$$For one-dimensional motion:
$$W = \int_{x_1}^{x_2} F(x)\,dx$$Graphical Interpretation
Work = Area under F-x curve
- Area above x-axis → Positive work
- Area below x-axis → Negative work
Work Done by Common Forces
1. Gravity
Falling through height h:
$$W_{gravity} = mgh \quad \text{(positive)}$$Rising through height h:
$$W_{gravity} = -mgh \quad \text{(negative)}$$2. Spring Force
For a spring compressed or extended by $x$ from natural length:
$$W_{spring} = -\frac{1}{2}kx^2$$This is always negative because spring force opposes the displacement.
Work done BY external force to stretch/compress:
$$W_{external} = +\frac{1}{2}kx^2$$3. Friction
For kinetic friction over displacement $s$:
$$W_{friction} = -\mu_k mg \cdot s = -f_k \cdot s$$Always negative because friction always opposes motion.
4. Normal Force
$$W_{normal} = 0$$The normal force is perpendicular to the surface, and for motion along the surface, it does no work.
Summary Table
| Force | Work Formula | Sign | Notes |
|---|---|---|---|
| Gravity (down) | $mgh$ | + | Path independent |
| Gravity (up) | $-mgh$ | - | Path independent |
| Spring | $-\frac{1}{2}kx^2$ | - | From natural length |
| Kinetic friction | $-f_k s$ | - | Path dependent |
| Normal force | $0$ | 0 | Perpendicular |
| Centripetal force | $0$ | 0 | Perpendicular |
Worked Example
Solution:
Work by applied force:
$$W_F = F \cdot s \cdot \cos37° = 30 \times 4 \times 0.8 = 96 \text{ J}$$Normal force:
$$N = mg - F\sin37° = 50 - 30(0.6) = 32 \text{ N}$$Work by friction:
$$W_f = -\mu_k N \cdot s = -0.2 \times 32 \times 4 = -25.6 \text{ J}$$Work by gravity and normal: Both = 0 (perpendicular)
Net work:
$$W_{net} = 96 + (-25.6) + 0 + 0 = \boxed{70.4 \text{ J}}$$
Practice Problems
A 10 N force acts at 60° to horizontal on a 2 kg object, displacing it 5 m horizontally. Find the work done.
Find the work done by a spring (k = 200 N/m) when stretched from x = 0.1 m to x = 0.3 m.
A 2 kg block slides 3 m down a rough incline (30°, $\mu_k = 0.3$). Calculate work done by each force.