Short on time? This page is the recap for the calculus-based course (separate from Physics 1). For derivations, graphs and worked examples, use the full study guide.
Recap
- Work is energy transferred into (W > 0) or out of (W < 0) a system by a force acting while its point of application moves. Scalar; joules (1 J = 1 N·m).
- Only the force component parallel to the displacement does work. A perpendicular component can turn the motion without changing K.
- Conservative forces (gravity, ideal springs): work depends only on start and end configurations; zero round a closed path; only these have a potential energy.
- Nonconservative forces (kinetic friction, air resistance): work depends on the path. Energy removed by friction ≈ f × path length.
- A system can be modelled as an object when the point of application and the centre of mass move the same distance.
Key relationships
| Relationship | When to use it |
|---|---|
| A · B = AB cos θ = A_xB_x + A_yB_y + A_zB_z | any dot product |
| W = F · d = Fd cos θ = F‖d | F‖ constant |
| W = ∫ₐᵇ F · dr; in 1D, W = ∫ F_x dx | force varies along the path |
| W = signed area under F‖–position graph | graph given |
| W_s = −½k(x_f² − x_i²) | ideal spring, x from relaxed length |
| W_net = ΔK = ½mv_f² − ½mv_i² | work–energy theorem, all forces |
Assumptions behind the numbers
- Point-object model unless the question says the system changes shape.
- g = 9.8 m/s²; a level surface means gravity and the normal force do no work on horizontal motion.
- Only mechanical energy is tracked. Friction’s lost energy becomes thermal energy and sound, which you only need to name.
Mistakes to avoid
- F × d with a varying force. Integrate or use the area.
- Wrong angle. θ is between F and d, not between F and the floor.
- Dropping signs. Friction on a sliding object does negative work.
- Work–energy theorem with one force. Add the work of every force.
- “Centripetal force does work.” It is perpendicular to the velocity: W = 0.
- Closed loop with friction gives zero. Only for conservative forces.
- Using the centre-of-mass displacement when the point of application does not move (pushing off a wall: W = 0).
Quick self-check
- F = (2.0, −5.0) N, d = (4.0, 1.0) m. Find W. (8.0 − 5.0 = 3.0 J)
- A 20 N force at 60° to a 3.0 m displacement. Find W. (20 × 3.0 × cos 60° = 30 J)
- F_x = 4.0x³ (N, x in m). Find the work from x = 0 to 1.0 m. ([x⁴]₀¹ = 1.0 J)
- A 0.50 kg ball goes from 2.0 m/s to 6.0 m/s. Find the net work. (½ × 0.50 × (36 − 4.0) = 8.0 J)
Next: practice questions.