Short on time? This page is the recap. For diagrams and worked examples, use the full study guide. This is the algebra-based course: no calculus is needed.
Recap
- Work is energy transferred into (+) or out of (−) a system by a force exerted over a distance. It is a scalar.
- Only the force component parallel to the displacement does work. A perpendicular force can turn an object without changing its kinetic energy.
- Work-energy theorem: the change in an object’s kinetic energy equals the net work done on it by all forces.
- Conservative forces (gravity, ideal springs): work depends only on the start and end positions, and is zero for a round trip. Only these have a potential energy.
- Nonconservative forces (friction, air resistance): work depends on the path.
- Energy dissipated by friction = friction force × path length. It becomes thermal energy and sound.
- Work is the area under a force–position graph; area below the axis is negative.
- A system can be treated as an object when its center of mass and the point where the force acts move the same distance.
Key relationships
| Relationship | Symbols and units | Use it to |
|---|---|---|
| W = F∥d = Fd cos θ | F in N, d in m, W in J | work by a constant force |
| ΔK = ΣW = W_net | J | link forces to a change in speed |
| W_g = −mgh (going up h) | J | work by gravity, any path |
| energy dissipated = F_f × path length | J | energy removed by friction |
| W = area under F_x–x graph | N × m = J | work by a changing force |
| d = mv₀² / (2F_f) | m | stopping distance; d ∝ v₀² |
Assumptions behind the numbers
- Forces are constant unless a graph is given.
- d is the displacement of the point where the force acts.
- Only mechanical energy is tracked; dissipated energy is described, not calculated in detail.
- g = 9.8 m/s².
Mistakes to avoid
- Using W = Fd for a force at an angle. Use F cos θ.
- Using sin θ instead of cos θ. θ is measured between the force and the displacement.
- Leaving out a sign. Friction on a sliding object does negative work.
- Using F_f = μmg when another force has a vertical component. Find F_N first.
- Using displacement for friction losses. Use path length.
- Forgetting a force when adding work. List every force from the free-body diagram.
- Saying gravity does more work on a longer ramp. It depends only on the height change.
Quick self-check
- A 50 N force pushes a box 4.0 m in the direction of the force. Work done? (200 J)
- A 2.0 kg ball rolls 3.0 m across a level floor. Work done by gravity? (0 J: gravity is perpendicular to the motion)
- A force rises steadily from 0 to 12 N over 0.30 m. Work done? (Triangle area: ½ × 0.30 × 12 = 1.8 J)
- Friction of 5.0 N acts on a box pushed 2.0 m out and 2.0 m back. Work done by friction? (−20 J)
Next: practice questions.