Understand that mechanical or electrical work done is equal to the energy transferred

1.7.2 Work

Learning Objective

In the Cambridge IGCSE 0625 syllabus students must understand that mechanical or electrical work done is equal to the energy transferred. In other words, work is the amount of energy moved from one system to another by a force (mechanical) or by an electric current (electrical).

Key Definitions

  • Work (W): the transfer of energy that occurs when a force acts on an object and moves it through a distance, or when an electric current moves charge through a potential difference.
  • Mechanical work: work done by a force that causes a displacement.
  • Electrical work: work done when charge is moved through a voltage.
  • Energy transferred: the quantity of energy that leaves one system and enters another; numerically, this equals the work done.
  • Scalar nature of work: work has magnitude only (no direction). It can be positive (energy added to the system), negative (energy removed), or zero (no energy transferred).

Fundamental Formulae

Mechanical work

Core (scalar) form – used when the force and displacement are parallel:

\$W = F\,s\$

General form – when the force makes an angle θ with the displacement:

\$W = F\,s\,\cos\theta\$

  • \$F\$ – magnitude of the force (N)
  • \$s\$ – displacement (m)
  • \$\theta\$ – angle between the force vector and the displacement vector

Electrical work

\$W = VQ \qquad\text{or}\qquad W = V I t\$

  • \$V\$ – potential difference (V)
  • \$Q\$ – charge transferred (C)
  • \$I\$ – current (A)
  • \$t\$ – time (s)

Because \$V\times Q\$ (or \$V I t\$) has the unit joule, the result is the amount of energy transferred by the circuit.

Relationship to Energy

The numerical value of work done on (or by) a system is exactly the amount of energy transferred to (or from) that system:

\$\boxed{\text{Work} = \text{Energy transferred}}\$

Consequently the joule (J) is the SI unit for both work and energy.

Units & Conversion Table

QuantitySymbolSI UnitDefinition
ForceFnewton (N)1 N = 1 kg·m·s⁻²
Displacementsmetre (m)
Work / EnergyWjoule (J)1 J = 1 N·m = 1 kg·m²·s⁻²
Potential DifferenceVvolt (V)1 V = 1 J·C⁻¹
ChargeQcoulomb (C)1 C = 1 A·s
CurrentIampere (A)
Timetsecond (s)

Sign of Work (Positive, Negative, Zero)

  • Positive work: the component of the force is in the same direction as the displacement (\$\cos\theta>0\$). Energy is added to the object. Example: a person pushes a sled forward.
  • Negative work: the force component opposes the displacement (\$\cos\theta<0\$). Energy is removed from the object. Example: the brakes on a bicycle exert a force opposite to the motion, doing negative work and converting kinetic energy into heat.
  • Zero work: the force is perpendicular to the displacement (\$\cos\theta=0\$) or there is no displacement. Example: the tension in a rope that only changes direction of a swinging pendulum does no work.

Worked Examples

  1. Mechanical work (parallel force)

    A student pushes a 15 kg crate across a smooth floor with a constant horizontal force of 60 N over a distance of 5 m.

    Solution:

    \$W = F\,s = 60\;\text{N}\times5\;\text{m}=300\;\text{J}\$

    The work is positive, so 300 J of kinetic energy is transferred to the crate.

  2. Mechanical work (angled force)

    A 10 N force pulls a sled up a 30° slope for 8 m.

    Solution:

    \$W = F\,s\,\cos30^\circ = 10\;\text{N}\times8\;\text{m}\times0.866 = 69.3\;\text{J}\$

    Positive work adds 69.3 J of energy to the sled.

  3. Electrical work

    A 12 V lamp draws 2 A for 3 minutes.

    Solution:

    \$t = 3\;\text{min}=180\;\text{s}\$

    \$W = V I t = 12\;\text{V}\times2\;\text{A}\times180\;\text{s}=4320\;\text{J}\$

    The battery transfers 4320 J of energy to the lamp.

  4. Negative work (braking)

    A car of mass 800 kg traveling at 20 m s⁻¹ is brought to rest by a constant braking force over 40 m.

    Solution (using \$W = \Delta KE\$):

    \$\Delta KE = \tfrac12 m vf^2 - \tfrac12 m vi^2 = 0 - \tfrac12(800)(20)^2 = -160\,000\;\text{J}\$

    The brakes do –160 kJ of work, removing that amount of kinetic energy from the car.

Common Misconceptions

  • “Work is done only if the force is in the direction of motion.”

    The correct statement is: work is done if the force has a component along the displacement. If the component is zero (force ⟂ displacement) no work is done.

  • “All energy transferred is called work.”

    Energy can also be transferred as heat, sound, light, etc. Only the part transferred by a force (mechanical) or by moving charge (electrical) is called work.

  • “V × I is work.”

    \$V I\$ is power (watts). Work (or energy) is obtained by multiplying power by the time the power acts: \$W = V I t\$.

  • “Work has a direction.”

    Work is a scalar; its sign indicates whether energy is added to (+) or removed from (–) the system.

Link to Power

Power is the rate at which work is done (or energy is transferred):

\$P = \frac{W}{t}\$

  • Mechanical: \$P = F v\$ when the force and velocity are parallel.
  • Electrical: \$P = V I\$.

Practice Questions

  1. Calculate the work done when a 10 N force pulls a sled 8 m up a slope that makes a 30° angle with the horizontal.
  2. A battery supplies 9 V and powers a device drawing 0.5 A for 2 hours. How much energy is transferred in kilojoules?
  3. Explain why a force applied perpendicular to the motion of an object does not transfer energy to the object.
  4. State whether the work done by gravity on a 2 kg mass falling 5 m is positive, negative or zero, and give its magnitude.

Suggested Diagram

Block being pushed across a horizontal surface showing force F, displacement s, and angle θ

Block being pushed across a horizontal surface: force \$F\$, displacement \$s\$, and angle \$\theta\$ between them.