A Labourer Moving a Loaded Cart — Work Done and Force Analysis

Work done by a force: W = F · d = F × d × cosθ

Where:

• W = work done (in joules, J)
• F = magnitude of applied force (in newtons, N)
• d = displacement of the cart (in metres, m)
• θ = angle between force and displacement

Special cases:

1. Force parallel to motion (θ = 0°): W = Fd × cos0° = Fd (maximum work)
2. Force at 90° to motion (θ = 90°): W = Fd × cos90° = 0 (no work done)
3. Force at 180° to motion (θ = 180°): W = Fd × cos180° = −Fd (negative work)

Example: A labourer pushes a cart with 100 N over 20 m horizontally. W = 100 × 20 × cos0° = 2000 J

While a labourer pushes a cart on a level road:

1. Applied force (horizontal): Does positive work = F × d
2. Friction force (opposing motion): Does negative work = −f × d
3. Normal force (vertical, upward): Work = 0 (perpendicular to displacement)
4. Gravity (vertically downward): Work = 0 (perpendicular to displacement on flat ground)

Net work done = Work by applied force − Work against friction = Fd − fd = (F − f) × d

By Work-Energy Theorem: Net work = ΔKE = ½mv² − ½mu²

If cart moves at constant speed: Net work = 0, so F = f (applied force = friction)

An important case: A porter (labourer) carrying a load on the head while walking horizontally.

• The force applied on the load = upward (to support it against gravity)
• The displacement of the load = horizontal
• Angle between force and displacement = 90°
• Work done by the porter on the load = F × d × cos90° = 0

This is a classic example of zero work in physics. The porter does no work in the physics sense on the load, even though they feel tired (their muscles are doing internal biological work).

Note: Work is done against gravity only when the load is lifted vertically.