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Physics · 09 / 12

Work of a force and energy balance

High school · intermediate · Worked example and 4 exercises with solutions, at your pace.

Need a foundation?Units: numbers have dimensionsEquations: keep the balance level
UnderstandSee an exampleGuided exercise3 progressive challenges

The principle, simply

For constant force over straight displacement, W=Fd cos α, with angle between force and displacement. Positive work favours motion; negative work opposes it. The work-energy theorem relates kinetic-energy change to total work on the chosen system.

W=F*d*cos(alpha)
THE PICTURE TO REMEMBERWork of a force and energy balance
  1. 01Force and path
  2. 02Angle
  3. 03Signed work
  4. 04All-force balance

WORKED EXAMPLE

From principle to reasoning

10 N parallel to 3 m gives 30 J. Perpendicular force gives zero work; opposite direction gives −30 J.

Solve a problem in this subject

  1. Draw the system

    Define the studied object, reference frame and positive axis. A force diagram shows external interactions acting on the system.

  2. Sort quantities

    Separate data and unknowns; convert units before substitution. 36 km/h = 10 m/s, so never insert 36 into a formula expecting SI speed.

  3. Write the model

    Choose a law matching assumptions: constant acceleration, thin lens, ohmic resistor. Isolate the unknown symbolically before rounding.

  4. Interpret

    Check dimensions and signs. Negative velocity gives direction on the chosen axis. Discuss scale and model limitations.

YOUR TURN

01 · Try before checking

A 4 N friction force opposes 5 m of motion. Find work.

One hint

Angle 180° gives cosine −1.

Reveal and understand the solution

W=−20 J. Without compensating work, kinetic energy decreases.

Compare method, conditions and answer; identify the exact part to revisit.

IMPROVE THROUGH PRACTICE

Three additional challenges with solutions

Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.

Reflect after practising
02 · Recognise and understandLink to this exercise

A 10 N force acts along a 3 m displacement. Find its work.

A hint to get started

W = Fd cos(θ).

Step-by-step solution
  1. Force and displacement are parallel: θ = 0.
  2. cos(0) = 1, giving W = 10 × 3.
  3. Work is +30 J: the force adds energy.

Which step did you miss? Retry without help, then explain why the method works.

03 · Apply the methodLink to this exercise

A 5 N friction force opposes a 4 m motion. Find its work.

A hint to get started

Force and displacement are at 180°.

Step-by-step solution
  1. cos(180°) = −1.
  2. W = 5 × 4 × (−1) = −20 J.
  3. Negative work reduces kinetic energy if other work is zero.

Which step did you miss? Retry without help, then explain why the method works.

04 · Go furtherLink to this exercise

A 2 kg body starts at rest. Total work is 25 J. Find final speed.

A hint to get started

Apply the work–kinetic-energy theorem.

Step-by-step solution
  1. Final kinetic energy is 25 J because initial energy is zero.
  2. mv²/2 = 25; with m = 2, v² = 25.
  3. The speed is 5 m/s.

Which step did you miss? Retry without help, then explain why the method works.

The trap to avoid

Nonzero force can have zero work. Check angle and assumptions.

Without notes, can you…

  • Explain the principle in your own words?
  • Rebuild the diagram and its connections?
  • Solve the exercise and justify the method?

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