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UY1: Work-Energy Theorem

Derive and use the work–energy theorem (net work equals change in kinetic energy) and connect it to power.

  • University Physics Year 1
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Learning objectives

  • Apply work–energy and momentum methods, then interpret the physical result.

The work–energy theorem is Newton’s 2nd law rewritten as an energy statement. It lets you relate forces and displacements directly to changes in speed without first solving for a(t) and v(t).

At a glance

  • Prerequisites: Concept of Work (UY1); Newton’s 2nd law (unbalanced force); vectors + dot product (Mathematics for Undergraduate Physics); basic integration (Integration Techniques)
  • You will learn: derive Wₙₑₜ = Δ K; apply it to speed/stopping-distance problems; compute power as an energy-transfer rate
  • Key result: Wₙₑₜ = Δ K = 1/2 m(v_f²-vᵢ²)
  • Common trap: summing the work of only one force instead of net work, or losing the sign (e.g. friction does negative work)

Setup (what we assume)

We model a particle of mass m moving under a net force vector Fₙₑₜ. From the previous lesson, the net work from point 1 to 2 is: Wₙₑₜ = ∫₁² vector Fₙₑₜ · d vector r.

Net work means ‘sum the work of all forces’

If multiple forces act, compute each work contribution along the path and add them: Wₙₑₜ = ∑ᵢ Wᵢ.

When this method is the best choice

Work-energy is usually fastest when:

  • the question asks for a speed or distance between two positions (not the full v(t)),
  • forces depend on position (springs, variable forces), or constraints make a(t) annoying,
  • you can compute work easily from geometry and dot products.

If the question asks for the time or needs detailed dynamics, Newton’s 2nd law is often the better tool.

1) The work–energy theorem

Define kinetic energy: K = 1/2 mv².

The work–energy theorem states: Wₙₑₜ = Δ K = K_f-Kᵢ.

Why this step matters

Writing Newton’s law in work form replaces time-based dynamics with state-to-state energy accounting. This is why you can solve speed or distance directly even when acceleration is not explicitly found.

Derivation (from Newton’s 2nd law)

Start from Newton’s 2nd law: vector Fₙₑₜ = m(d vector v)/dt.

Dot both sides with the displacement element d vector r: vector Fₙₑₜ · d vector r = m(d vector v)/dt · d vector r.

Use d vector r = vector v dt: m(d vector v)/dt · d vector r = m(d vector v)/dt · vector v dt = m vector v · d vector v.

Integrate from initial to final states: ∫₁² vector Fₙₑₜ · d vector r = m∫_(vector vᵢ)^(vector v_f) vector v · d vector v = (1/2)m(v_f²-vᵢ²).

Recognize the left side as Wₙₑₜ and the right side as Δ K.

How to use it (workflow)

  1. Choose your system and identify all forces that do work.
  2. Compute each work contribution along the displacement.
  3. Set Wₙₑₜ = Δ K and solve for the unknown (often v_f or a distance).

This method is especially useful when the force varies with position, or when time is not asked for.

Common pitfalls

  • Computing work from only one force when multiple forces act.
  • Dropping signs for resistive work (friction/drag usually contributes negative work on the moving object).
  • Treating centripetal force as doing work in uniform circular motion when it is perpendicular to displacement.

Mini-example: variable force (spring stopping distance)

A mass m = 0.50 kg slides on a frictionless surface with speed vᵢ = 4.0 m s⁻¹ and compresses a spring (k = 200 N m⁻¹) until it momentarily stops. Find the maximum compression x.

Work done by the spring: W = ∫₀^x (-kx') dx' = -(1/2)kx².

Work-energy theorem with v_f = 0: -(1/2)kx² = Δ K = 0-(1/2)mvᵢ². So x = vᵢ square root of (m/k) = 4.0 square root of (0.50/200) = 0.20 m.

Checks: x decreases if k increases; units: square root of (m/k) has units of seconds, so vᵢ square root of (m/k) is metres.

2) Power (rate of doing work)

Average power: Pₐᵥ = (Δ W)/(Δ t).

Instantaneous power: P = dW/dt = vector F · (d vector r)/dt = vector F · vector v.

Units: W = J s⁻¹. (And 1 hp ≈ 746 W.)

Worked example

Example: stopping distance under a constant braking force

Given: a car of mass m = 900 kg travels at vᵢ = 22 m s⁻¹ and comes to rest under a constant braking force of magnitude F = 3600 N opposite the motion.

Find: the stopping distance d.

Working: braking does negative work, so Wₙₑₜ = -Fd. Also Δ K = 0-(1/2)mvᵢ². -Fd = -(1/2)mvᵢ² ⇒ d = ((1/2)mvᵢ²)/F. Numerically: d = 0.5(900)(22²)/3600 ≈ 6.1 × 10¹ m.

Answer + check: d ≈ 61 m; units are metres, and a larger force gives a shorter stopping distance.

Practice set (with hints + answers)

  1. A 2.0 kg object speeds up from 3.0 to 7.0 m s⁻¹. Find the net work done.
  2. A 0.20 kg ball moving at 10 m s⁻¹ comes to rest after 0.50 m in sand. Estimate the average resistive force magnitude.
  3. A 5.0 kg block starts from rest on a frictionless surface. A constant horizontal force 12 N acts over 4.0 m. Find the final speed.
  4. A motor delivers 600 W to lift a 15 kg mass at constant speed. Find the lifting speed.
  5. In uniform circular motion at constant speed, what is the net work done by the (ideal) centripetal force over one full circle?

Hints

  1. Use W = Δ K = -Fₐᵥd.
  2. Work done is W = Fd; then W = Δ K.
  3. Use P = vector F · vector v = mgv (force and velocity are collinear).
  4. Use “force perpendicular to motion” ⇒ zero work.

Answers

  1. Wₙₑₜ = 1/2 m(v_f²-vᵢ²) = 40 J.
  2. Fₐᵥ = ((1/2)mv²)/d = 20 N.
  3. v_f = square root of (2Fd/m) ≈ 4.38 m s⁻¹.
  4. v = P/mg ≈ 4.1 m s⁻¹.
  5. W = 0.

Summary + next steps

  • Net work equals change in kinetic energy: Wₙₑₜ = Δ K.
  • Kinetic energy is K = (1/2)mv² (scalar, units J).
  • The theorem is often faster than solving F = ma when time isn’t asked for.
  • Power is the rate of work: P = vector F · vector v.

Next: Potential Energy & Conservative Forces Previous: Concept of Work Back To Mechanics (UY1)