What is work? (constant force)
In physics, work is done only when a force makes something move, and the motion has a part along the force.
W = F · s = F s cos θ, where θ is the angle between force F and displacement s.
- θ = 0°: W = Fs (largest, positive).
- θ = 90°: W = 0. Example: carrying a bag while walking on level ground, the lifting force is up, motion is sideways.
- 90° < θ ≤ 180°: W is negative. Example: friction and brakes.
Work is a scalar (it has no direction). Unit: joule (J) = 1 N × 1 m. Dimensions: [ML²T⁻²].
Work by a variable force
If the force changes with position, we cannot just multiply. We break the path into tiny pieces Δx. In each piece the force is almost constant, so the small work is F(x) Δx.
Add all pieces: W = ∫ F(x) dx from x₁ to x₂. On a graph, this is the area under the F–x curve. Area above the x-axis counts positive, below counts negative.
Example: a spring force F = kx gives a triangle, so W = ½kx².
Kinetic energy
Kinetic energy is the energy a body has because it is moving: K = ½mv². It is a scalar and is never negative.
Double the speed → K becomes 4 times. Link with momentum p = mv: K = p²/2m.
Work–energy theorem and its proof
Statement: the work done by the net force on a body equals the change in its kinetic energy: W_net = K_f − K_i.
Proof for a constant force:
- From kinematics: v² − u² = 2as.
- Multiply both sides by m/2: ½mv² − ½mu² = mas.
- But ma = F (Newton's second law), so ½mv² − ½mu² = Fs = W.
Proof for a variable force: F = m dv/dt = m (dv/dx)(dx/dt) = m v dv/dx. So F dx = m v dv. Integrate: ∫F dx = ½mv² − ½mu².
It uses the net work (all forces together).
Power
Power is the rate of doing work.
- Average power: P = W/t.
- Instantaneous power: P = dW/dt = F · v = F v cos θ.
Unit: watt (W) = 1 J/s. 1 horsepower (hp) = 746 W. 1 kilowatt-hour (kWh) = 3.6 × 10⁶ J is a unit of energy, not power.
Try it at home
Take a school bag. Pull it 2 m on the floor with a string held low, then with the string held high. Which felt easier to move forward? Now run up a flight of stairs, then walk up. The work (your weight × height) is the same, but running needs more power. Time yourself: P = mgh ÷ t.
Exam corner
CBSE often asks: define work with θ cases (1–2 marks), prove the work–energy theorem (3 marks), numericals on F–x graph area, power = F·v, and ratios of K for given momentum. This unit shares 17 marks with units IV–VI.
Key formulas and definitions
- W = F · s = F s cos θ
- W = ∫ F(x) dx = area under F–x graph
- K = ½mv² = p²/2m
- W_net = K_f − K_i
- P = W/t, P = F · v = F v cos θ
- 1 hp = 746 W, 1 kWh = 3.6 × 10⁶ J
Worked examples
1. A 5 N force moves a box 3 m along its own direction. Find the work.
Step 1: θ = 0°, cos 0° = 1. Step 2: W = F s = 5 × 3. Answer: 15 J.
2. A child pulls a toy cart with 20 N at 60° to the ground over 5 m. Find the work.
Step 1: W = F s cos θ. Step 2: cos 60° = 0.5. Step 3: W = 20 × 5 × 0.5 = 50 J.
3. Find the kinetic energy of a 60 kg runner at 5 m/s.
Step 1: K = ½mv². Step 2: K = ½ × 60 × 25 = 750 J.
4. A force F = (3î + 4ĵ) N moves a body by s = (2î + ĵ) m. Find the work.
Step 1: W = F · s = FₓSₓ + F_yS_y. Step 2: W = 3 × 2 + 4 × 1 = 10 J.
5. A force F = 2x N (x in m) acts from x = 1 m to x = 3 m. Find the work.
Step 1: W = ∫ 2x dx from 1 to 3. Step 2: = [x²] from 1 to 3 = 9 − 1. Answer: 8 J. (It is also the area of the trapezium under the line: ½ × (2 + 6) × 2 = 8 J.)
6. A 2 kg ball moving at 3 m/s is pushed by a net force of 8 N over 4 m in its direction. Find its final speed.
Step 1: W = 8 × 4 = 32 J. Step 2: K_i = ½ × 2 × 9 = 9 J. Step 3: K_f = 9 + 32 = 41 J. Step 4: ½ × 2 × v² = 41 → v² = 41 → v ≈ 6.4 m/s.
7. A car of 1000 kg moving at 20 m/s is brought to rest in 40 m. Find the average braking force.
Step 1: ΔK = 0 − ½ × 1000 × 400 = −2 × 10⁵ J. Step 2: W = −F × 40. Step 3: F = 2 × 10⁵ / 40 = 5000 N.
8. A pump lifts 600 kg of water to 20 m in 2 minutes. Find its power (g = 10 m/s²).
Step 1: W = mgh = 600 × 10 × 20 = 1.2 × 10⁵ J. Step 2: t = 120 s. Step 3: P = 1.2 × 10⁵ / 120 = 1000 W = 1 kW (≈ 1.34 hp).
Common mistakes
- Using W = Fs when the force is at an angle. Always use F s cos θ, or the dot product.
- Thinking a person holding a heavy bag still does work on it. No displacement → no work (in the physics sense), though muscles get tired.
- Using only one force in the work–energy theorem. The theorem uses the net work of all forces.
- Mixing up kWh and kW. kW is power; kWh is energy (what the electricity bill counts).