Study Guides/Physics/A Labourer Moving a Loaded Cart — Work Done in Physics
Study Guide · Physics

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

When a labourer moves a loaded cart, the work done is calculated using the formula W = F × d × cosθ, where F is the applied force, d is the displacement, and θ is the angle between the force and the direction of motion. If the force is applied horizontally and displacement is horizontal, maximum work is done (θ = 0°, W = Fd). If the force is perpendicular to displacement, work done is zero.

Question (Click to Flip)

How do you calculate work done by a labourer moving a cart?

Answer

Work done = F × d × cosθ, where F is the applied force, d is the displacement, and θ is the angle between force and displacement. For a horizontal push on a level road (θ = 0°), W = F × d.

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Key Facts

Work done = F × d × cosθ, where θ is the angle between force and displacement.

If force is parallel to displacement (θ = 0°), work done is maximum: W = Fd.

If force is perpendicular to displacement (θ = 90°), work done is zero.

A porter carrying a load horizontally does zero work on the load (force is vertical, displacement is horizontal).

SI unit of work is the joule (J); 1 J = 1 N × 1 m.

Net work done = change in kinetic energy (work-energy theorem).

Work done against friction is converted to heat energy.

Work Done Formula for Moving a Loaded Cart

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

Normal Force and Work Done by Gravity

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)

When a Porter Carries a Load on the Head — Zero Work

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.

Questions and Answers

How do you calculate work done by a labourer moving a cart?+

Work done = F × d × cosθ, where F is the applied force, d is the displacement, and θ is the angle between force and displacement. For a horizontal push on a level road (θ = 0°), W = F × d.

A labourer carries a load on his head and walks horizontally — how much work is done?+

Zero work is done on the load. The force applied by the labourer is vertical (upward), while the displacement is horizontal. Since the angle between force and displacement is 90°, W = F × d × cos90° = 0.

What is the work done against friction when a cart is moved?+

Work done against friction = friction force × displacement = f × d = μmg × d, where μ is the coefficient of friction, m is the mass, and g is gravitational acceleration. This work is dissipated as heat.

If a labourer pushes a cart with 200 N over 15 m, what is the work done?+

Work done = F × d × cosθ = 200 × 15 × cos0° = 200 × 15 × 1 = 3000 J (3 kJ), assuming the force is applied horizontally in the direction of motion.

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