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Dynamics · Chapter 4 of 10 · Intermediate

Work and Energy

Newton's law connects force to acceleration at an instant. Work and energy connect force to speed over a distance, often skipping the acceleration entirely.

Readiness check

This chapter recasts Chapter 3 in scalar form. Tick only what you can do closed-notes.

Write ΣF = ma and draw a free body.
Compute a dot product of force and displacement.
Integrate a force over a distance.
Recall the spring force F = kx.
Track signs of energy gained and lost.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview forces and friction in Chapter 3.

3 or more weak itemsRevisit work and energy basics in virtual work before continuing.

The core idea

The work done on a particle equals its change in kinetic energy; when the only forces are conservative, the total of kinetic and potential energy stays constant.

U₁₋₂ = T₂ − T₁T = ½mv²T₁ + V₁ = T₂ + V₂

Energy methods trade the vector equation ΣF = ma for a single scalar equation. When a problem asks for speed as a function of position, and time and acceleration are not wanted, work-energy is far quicker than integrating Newton's law. If every force is conservative (gravity, springs), the bookkeeping collapses to "energy in equals energy out."

The skill works when: you want a speed-distance relation and can total the work of every force, or recognise a conservative system.

The skill breaks down when: friction is forgotten in the energy balance, or energy methods are used to find an acceleration or a time directly.

The concept. A force acting through a displacement does work, and that work appears as a change in kinetic energy. Speed at the end follows from speed at the start plus the net work, with no need for acceleration or time.

The skills, taught in order

Energy methods are a small toolkit: define work, total it, equate to the change in kinetic energy, and use potential energy when forces are conservative. Five skills cover it.

4.1 The work of a force

Work is the force times the displacement in its own direction: U = ∫F·dr. Only the component of force along the motion does work; a force perpendicular to the path (like a normal force) does none. Work is a scalar, measured in joules, and can be positive or negative.

4.2 Work of common forces

A few forms recur in almost every problem.

Force	Work done	Sign
Weight (drop h)	U = −mgΔy = +mgh falling	+ down, − up
Spring	U = −½k(x₂² − x₁²)	− when stretched or compressed further
Friction	U = −F_f·d	always negative
Normal force	U = 0	perpendicular to motion

4.3 Kinetic energy and the work-energy theorem

Kinetic energy is T = ½mv². The theorem states U₁₋₂ = T₂ − T₁: the total work of all forces equals the change in kinetic energy. This single scalar equation replaces the vector ΣF = ma when speed and distance are the unknowns.

4.4 Potential energy and conservation

Conservative forces store energy: gravitational V_g = mgy and elastic V_e = ½kx². For a system with only conservative forces, T₁ + V₁ = T₂ + V₂: mechanical energy is conserved. Any friction or applied work is added as a separate non-conservative term.

4.5 Power and efficiency

Power is the rate of doing work, P = dU/dt = F·v, measured in watts. Mechanical efficiency η is the ratio of useful power out to power in, always less than one because friction dissipates some input. Power decides motor and engine sizing.

Engineering connection: braking distances, roller-coaster and ride design, spring and impact energy, and the power ratings of motors and engines.

Worked example 1: speed at the bottom of a chute

A 20 kg package is released from rest at the top of a 4 m chute inclined at 30°, with kinetic friction μ_k = 0.2. Use work-energy to find its speed at the bottom.

Figure 1. Gravity does positive work as the package drops 2 m; friction does negative work over the 4 m slope. The net work becomes kinetic energy at the bottom.

ProblemFind the speed at the bottom of the chute in Figure 1 using energy.
Given / findm = 20 kg, d = 4 m, θ = 30°, μ_k = 0.2, from rest. Find v.
AssumptionsParticle, constant kinetic friction, normal force does no work.
ModelTotal the work of gravity and friction, set it equal to the kinetic energy gained (starts from rest, so T₁ = 0).
EquationsU_grav = mgh, h = d sinθ U_fric = −μ_kmg cosθ · d U_grav + U_fric = ½mv²
Solveh = 4 sin30° = 2.0 m, so U_grav = 20 × 9.81 × 2 = 392.4 J. Friction force = 0.2 × 20 × 9.81 × cos30° = 33.98 N, so U_fric = −33.98 × 4 = −135.9 J. Net work = 256.5 J = ½(20)v², giving v = √(2 × 256.5/20) = 5.06 m/s.
CheckWithout friction the speed would be √(2gh) = 6.26 m/s; friction has reduced it, as it must. The normal force never entered, because it does no work, which is exactly why energy is quicker here.
ConclusionOne scalar equation gave the answer with no acceleration or time. Energy methods shine whenever the question links speed to distance.

Result. Speed at the bottom v = 5.06 m/s (versus 6.26 m/s frictionless).

Worked example 2: ramp into a spring

A 2 kg block slides from rest down a frictionless ramp through a height of 1.5 m, then strikes a horizontal spring of stiffness k = 800 N/m. Find the maximum compression of the spring.

Figure 2. On the frictionless ramp, gravitational potential energy converts entirely to kinetic energy, then to elastic energy in the spring at maximum compression, where the block is momentarily at rest.

ProblemFind the maximum spring compression for the block in Figure 2.
Given / findm = 2 kg, h = 1.5 m, k = 800 N/m, frictionless, from rest. Find x_max.
AssumptionsNo friction, spring is horizontal so no height change during compression, block at rest at maximum compression.
ModelConservation of energy from the release point to maximum compression: all gravitational potential energy becomes elastic potential energy.
EquationsT₁ + V₁ = T₂ + V₂ mgh = ½kx² x = √(2mgh/k)
SolveAt release and at maximum compression the block is at rest, so kinetic energy is zero at both ends. mgh = 2 × 9.81 × 1.5 = 29.4 J = ½(800)x². So x = √(2 × 29.4/800) = √0.0736 = 0.271 m. The peak spring force is kx = 800 × 0.271 = 217 N.
CheckThe block reaches the bottom at √(2gh) = 5.42 m/s, and ½mv² = 29.4 J matches mgh, confirming the energy bookkeeping. The compression scales as √h, so a four-times-higher drop only doubles it.
ConclusionEnergy conservation handled two conversions in one line. The intermediate speed was never needed, though it cross-checks the result.

Result. Maximum compression x = 0.271 m, peak spring force 217 N.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Omitting friction from energy	Speed comes out too high	"Did I subtract the friction work?"	Friction always removes energy; include U = −F_fd.
Counting normal-force work	An extra work term appears	"Is this force along the motion?"	Perpendicular forces do no work; drop them.
Using energy for acceleration	Stuck trying to get a or t	"Does the question ask speed vs distance?"	Energy gives speed-distance; use ΣF = ma for a or t.
Double-counting gravity	Weight in both work and potential energy	"Did I pick work or potential energy for gravity?"	Use one or the other, not both, in a single balance.

Practice ladder

Level 1 · Direct skill

A 1500 kg car moving at 20 m/s brakes to rest. How much kinetic energy must the brakes remove?

Show answer

T = ½mv² = ½ × 1500 × 20² = 300 000 J = 300 kJ. All of it converts to heat in the brakes, which is why repeated hard braking overheats them.

Level 2 · Mixed concept

For the Worked Example 1 chute, what slope angle would let the package reach the bottom with the same speed as a frictionless 30° chute?

Show answer

Set g(sinθ − μ_kcosθ)d equal to the frictionless g sin30° d ... this needs a steeper angle so the extra gravity work offsets friction. Solving sinθ − 0.2cosθ = 0.5 gives θ ≈ 38.7°. Friction must be paid for with more height.

Level 3 · Independent problem

A 30 kg child on a 4 m swing (treated as a pendulum) starts from rest at 40° from vertical. Find the speed at the bottom, ignoring friction.

Show answer

Height drop h = L(1 − cos40°) = 4(1 − 0.766) = 0.936 m. Then v = √(2gh) = √(2 × 9.81 × 0.936) = 4.29 m/s. Energy conservation turns the geometry of the drop straight into speed.

Level 4 · Transfer to real engineering

Pick a real energy conversion (a braking vehicle, a bouncing ball, a wind-up toy). Set up a work-energy or conservation balance, identify where energy is lost, and estimate a speed or distance.

What good work looks like

Every work or energy term listed with its sign, conservative and non-conservative parts separated, and a result checked against a frictionless limit.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I included every work term with the right sign before I equate to ΔT."

"Give me five problems; I will decide whether energy or ΣF = ma is the faster route."

"Solve for the final speed." Building the energy balance is the skill.

"Is energy conserved here?" Spotting the non-conservative forces is the judgment being trained.

Portfolio task

Analyse one real system with an energy balance, explicitly separating conservative storage from friction losses, and end with a speed or distance plus an efficiency estimate.

Must include: a labelled energy accounting, the work-energy theorem or conservation applied, and a frictionless-limit check.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. State the work-energy theorem.

U₁₋₂ = T₂ − T₁: total work equals the change in kinetic energy.

2. Give kinetic and the two potential energies.

T = ½mv²; gravitational V = mgy; elastic V = ½kx².

3. How much work does a normal force do?

None; it is perpendicular to the motion.

4. When is mechanical energy conserved?

When all forces doing work are conservative (no friction or applied work).

5. Define power and efficiency.

Power P = F·v (rate of work); efficiency η = useful power out / power in.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-solve the chute by energy from a blank page.

+3 daysOne conservation problem (pendulum or spring).

+7 daysContrast energy with the momentum method of Chapter 5.

+30 daysCarry energy into rigid bodies in Chapter 9.

Textbook mapping

Item	Mapping
Primary source	Meriam and Kraige, Engineering Mechanics: Dynamics (7th ed), Chapter 3, Section B (Work and Energy)
Cross-reference	Hibbeler, Dynamics, Ch. 14 · Beer and Johnston, Ch. 13
Core topics	4.1 Work of a force · 4.2 Work of common forces · 4.3 Work-energy theorem · 4.4 Potential energy and conservation · 4.5 Power and efficiency
Engineering connection	Braking distances, ride design, spring and impact energy, and motor power sizing.
Read next	Chapter 5: Impulse, Momentum, and Impact.