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Work & Kinetic Energy

Key idea: Define work using the force component along displacement, derive kinetic energy from work and constant-acceleration equations, and apply the work–energy theorem.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Define work and derive and apply the kinetic-energy relationship.
  • Derive Eₖ = ½mv² from the definition of work done by a force and the uniformly accelerated motion equations.

1. Definitions (Must Know)

A. Work done, W (J)

Work done, W, is energy transferred by a force acting through a displacement in the direction of the force.

For a constant force at angle θ to the displacement:

W = Fs cos θ

Work is positive when the force component is along the displacement, negative when it is opposite, and zero when force and displacement are perpendicular.

B. Kinetic energy, Eₖ (J)

Kinetic energy, Eₖ, is energy a body has due to motion: Eₖ = (1/2)mv²

C. Work–energy theorem

The net work done on a body equals the change in kinetic energy: Wₙₑₜ = Δ Eₖ

2. Key Ideas (What Earns Marks)

  • Use the resultant force when linking net work to acceleration; use an individual force when asked for the work done by that force.
  • For straight-line motion with constant mass:
    • Wₙₑₜ = Fs
    • F = ma
    • v² = u² + 2as (uniform acceleration)
  • Combine them to get:
    • Wₙₑₜ = (1/2)m(v²-u²) = Δ Eₖ
    • so Eₖ = (1/2)mv²
  • Sign matters:
    • if the resultant force is opposite the displacement, work is negative and kinetic energy decreases.
Work is a scalar with a sign

Choose the displacement direction, then use the force component along it. Do not attach a vector arrow to work or kinetic energy.

3. Detailed Explanations

A. Derivation of Eₖ = (1/2)mv² (exam-ready)

Assume straight-line motion under a constant resultant force (so acceleration is constant).

Wₙₑₜ = Fs; = mas; Using v² = u² + 2as ⇒ as = (v²-u²)/2; Wₙₑₜ = m((v²-u²)/2); = (1/2)m(v²-u²)

So,

Wₙₑₜ = Δ Eₖ = (1/2)m(v²-u²)

This implies the kinetic energy of a body moving at speed v is:

Eₖ = (1/2)mv²

B. How to use the work–energy theorem

  1. Write Wₙₑₜ = Δ Eₖ.
  2. Express Wₙₑₜ using the forces that do work (often Fs along the motion).
  3. Solve for the unknown (speed, distance, force, etc).

C. Visual: negative work (braking / friction)

If the resultant force is opposite the displacement, the force–displacement graph lies below the axis and the work done is negative (kinetic energy decreases).

Resultant force vs displacement (constant braking force)

A constant resistive force acts opposite to the motion. The rectangular area under the line is the (negative) work done.

Scroll across the graph to read all labels.

A constant resistive force acts opposite to the motion. The rectangular area under the line is the (negative) work done.A constant resistive force acts opposite to the motion. The rectangular area under the line is the (negative) work done.
Example: F = -4800 N over s = 50 m gives W = Fs = -2.4 × 10⁵ J, so Δ Eₖ is negative.
Open full-size graph
View figure data
Values for Resultant force vs displacement (constant braking force)
Displacement, s (m)Braking force
0-4800
50-4800

4. Common Mistakes

  • Using the applied force instead of the resultant force in Wₙₑₜ.
  • Using distance travelled when the displacement in the force direction is required.
  • Forgetting unit conversions (e.g. cm → m, kN → N).
  • Using a SUVAT equation without checking the condition of constant acceleration.

5. Exam Tips

  • Use Wₙₑₜ = (1/2)m(v²-u²) when the question gives distance and asks for speed (or vice versa).
  • If the object stops, set v = 0 and solve for the stopping distance or braking force.
  • Keep signs consistent: work by a resistive force is negative.

6. Worked Examples

Modelled example 1

Net work from a change in speed

Core

Problem

A 2.0 kg object speeds up from 3.0 m s⁻¹ to 7.0 m s⁻¹. Find the net work done on it.
Study the worked solution

  1. Choose the work–energy relation

    Method

    Wₙₑₜ = Δ Eₖ.

    Reason

    The question asks for net work and supplies initial and final speeds, so no force or time is needed.

    Working

    Wₙₑₜ = E_(k,f)-E_(k,i)

  2. Calculate the energy change

    Method

    Wₙₑₜ = 40 J.

    Reason

    Kinetic energy depends on speed squared.

    Working

    Wₙₑₜ = (1/2)(2.0)(7.0²-3.0²) = 40 J

Guided practice 2

Speed from work done

About 5 min

Problem

A constant resultant force of 6.0 N acts on a 3.0 kg trolley initially at rest over 5.0 m. Find its final speed.

Try this before viewing the solution

Unit: m s^-1

Hints

Hint 1: transfer work into kinetic energy
Because the trolley starts from rest, (6.0)(5.0) = (1/2)(3.0)v².
View solution step by step

  1. Calculate net work

    Method

    Wₙₑₜ = 30 J.

    Reason

    The resultant force is along the displacement and constant.

    Working

    Wₙₑₜ = Fs = (6.0)(5.0) = 30 J

  2. Set the kinetic-energy gain

    Method

    30 = (1/2)(3.0)v².

    Reason

    The initial kinetic energy is zero.

    Working

    Wₙₑₜ = Δ Eₖ = (1/2)mv²

  3. Calculate final speed

    Method

    v = 4.47 m s⁻¹.

    Reason

    Speed is the positive square root of 2W/m.

    Working

    v = square root of (2(30)/3.0) = 4.47 m s⁻¹

Common misconception 3

Braking force from stopping distance

Find and correct the mistake

Learner claim

A 1200 kg car slows from 20 m s⁻¹ to rest over 50 m. A learner writes the braking work as + Fs because force magnitude and distance are positive. Diagnose the sign and find the constant braking-force magnitude.

Try this before viewing the solution

Unit: N

View solution step by step

  1. Determine the energy change

    Method

    Δ Eₖ = -2.4 × 10⁵ J.

    Reason

    The car loses all its initial kinetic energy.

    Working

    Δ Eₖ = (1/2)(1200)(0²-20²) = -2.4 × 10⁵ J

  2. Give braking work its sign

    Method

    W_brake = -F(50).

    Reason

    The force direction is opposite to the car’s displacement.

    Working

    W = Fs cos 180° = -Fs

  3. Find the force magnitude

    Method

    F = 4.8 × 10³ N.

    Reason

    Negative braking work equals the negative kinetic-energy change.

    Working

    -50F = -2.4 × 10⁵ ⇒ F = 4.8 × 10³ N

Examiner practice 4

Stopping distance from a constant resistive force

4 marks

Examination question

A 900 kg car travels at 25 m s⁻¹. Its brakes provide a constant resistive force of magnitude 5.0 × 10³ N. Ignoring other resistance, estimate the stopping distance. [4 marks]

Try this before viewing the solution

Unit: m

View solution step by step

  1. Write the signed work–energy equation

    2 marks

    Method

    -Fs = 0-(1/2)mu².

    Reason

    The car stops and the resistive force acts opposite its displacement.

    Working

    -Fs = Δ Eₖ = (1/2)m(0²-u²)

  2. Rearrange for distance

    1 mark

    Method

    s = mu²/(2F).

    Reason

    Both sides of the signed equation are negative, leaving a positive distance.

    Working

    s = (1/2)mu²/F

  3. Calculate the stopping distance

    1 mark

    Method

    s = 56.3 m, about 56 m.

    Reason

    Substitute the initial speed and constant resistive-force magnitude.

    Working

    s = ((1/2)(900)(25²))/(5.0 × 10³) = 56.25 m

Challenge 5

Given frictional force on a horizontal surface

Minimal support

Independent transfer

A 4.0 kg block slides on a rough horizontal surface. A constant 11.8 N frictional force opposes its motion, and it comes to rest after 2.5 m. Find the work done by friction and the block’s initial speed.

Try this before viewing the solution

Unit: J
Unit: m s^-1

Hints

Hint 1: connect the negative work to stopping
Set -fs = 0-(1/2)mu² after calculating the frictional work.
View solution step by step

  1. Calculate frictional work

    Method

    W_f = -29.5 J.

    Reason

    Friction acts at 180° to displacement.

    Working

    W_f = -fs = -(11.8)(2.5) = -29.5 J

  2. Relate work to the energy loss

    Method

    -29.5 = -(1/2)(4.0)u².

    Reason

    The block’s final kinetic energy is zero and friction is the only force doing work.

    Working

    Wₙₑₜ = 0-(1/2)mu²

  3. Find the initial speed

    Method

    u = 3.84 m s⁻¹.

    Reason

    Speed is the positive root of twice the removed energy divided by mass.

    Working

    u = square root of (2(29.5)/4.0) = 3.84 m s⁻¹

7. Mind Stretchers

Mind stretcher 1: Derive a stopping distance expressionExtension

Show that if a constant resistive force of magnitude F stops a mass m moving initially at speed u, the stopping distance is s = mu²/2F.

Show Answer

Using Wₙₑₜ = Δ Eₖ and v = 0: -Fs = (1/2)m(0²-u²); Fs = (1/2)mu²; s = mu²/2F

Mind stretcher 2: When does the work–energy method beat SUVAT?Extension

Give one example situation where Wₙₑₜ = Δ Eₖ is safer/faster than SUVAT, and explain why.

Show Answer

Example: “a car stops over a known distance with an unknown braking force” (or “a block slides to rest under friction over a distance”).

Energy is faster because it relates forces and distance directly: Wₙₑₜ = -Fs = Δ Eₖ without needing time. SUVAT would require finding/using time or constant acceleration assumptions explicitly; the energy method is direct and often less algebra.

Mind stretcher 3: Optional (Enrichment)Extension

A. Why you can feel tired when W = 0

In Physics, work done on an object depends on displacement in the direction of the force. So an object can have zero work done on it even though your body uses energy.

Show Explanation

Scenario A (pushing a wall/tree): The object does not move, so displacement is zero and the work done on the object is zero. But your muscles still use chemical energy to maintain tension, and much of that energy ends up as internal energy (heat) in your body.

Scenario B (carrying a load horizontally at constant speed): Your force on the load is mainly upward (to balance its weight) while the displacement is horizontal, so your force does zero work on the load. You can still feel tired because your muscles continuously contract to support the load and keep your posture stable.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027