14.1 Why Energy Matters in Physics
Energy is one of the most important concepts in all of science. It appears in every branch of physics — mechanics, thermodynamics, electromagnetism, quantum mechanics, and relativity. It is conserved: the total energy of an isolated system never changes. It takes many forms — kinetic, potential, thermal, chemical, nuclear, electromagnetic — and converts among them. These two features — conservation and multiple forms — make energy an extraordinarily powerful analytical tool.
In mechanics, the energy perspective provides a completely different way to analyze motion from the force perspective. Instead of asking “what forces act on this object and how do they produce acceleration?” Newton’s approach, the energy approach asks “how much work was done on this object and how did that change its kinetic energy?” Both approaches give the same answer, but each is more convenient in different situations.
The energy approach is especially powerful when forces are complicated, variable, or unknown in detail — as in collisions, spring systems, and problems where objects travel along curves. The kinematic and force methods require knowing force at every instant; energy methods require only knowing the total work done, regardless of the path taken.
Albert Einstein, paraphrase of the first law of thermodynamics
14.2 What Physicists Mean by Work
In everyday language, “work” means any effort or activity. In physics, work has a precise definition that does not always match everyday usage. Physical work is done on an object when a force acts on it and the object moves in the direction of the force, or has a component of displacement in the direction of the force.
Work is a scalar quantity — it has magnitude and sign but no direction. It is measured in joules (J), where 1 J = 1 N·m = 1 kg·m²/s². Work transfers energy from one object or system to another.
The physical work done equals the force applied times the displacement in the direction of the force. If the force and displacement are parallel, W = Fd. If they are at an angle, only the component of force along the displacement does work.
14.3 Conditions Required for Work
Three conditions must all be met for a force to do work on an object:
- A force must act on the object.
- The object must undergo a displacement; it must move.
- There must be a component of the force along the direction of displacement; the force cannot be purely perpendicular to the motion.
If any of these conditions is not satisfied, the force does zero work on the object. This leads to some counterintuitive results: gravity does no work on an object moving horizontally, the normal force does no work on an object sliding on a horizontal surface, and a centripetal force does no work because it is always perpendicular to velocity.
| Situation | Force | Displacement | Work Done? | Why |
|---|---|---|---|---|
| Pushing a box forward | Forward | Forward | Yes | Force parallel to displacement |
| Carrying a box horizontally | Upward | Horizontal | No | Force perpendicular to displacement |
| Lifting a box upward | Upward | Upward | Yes | Force parallel to displacement |
| Ball in circular orbit | Centripetal, inward | Tangential | No | Force always perpendicular to velocity |
| Friction on sliding box | Backward | Forward | Yes, negative | Force anti-parallel to displacement |
| Holding box stationary | Upward | Zero | No | No displacement |
14.4 Positive, Negative, and Zero Work
Work can be positive, negative, or zero, depending on the angle between the force and the displacement.
- Positive work (θ < 90°): the force has a component in the direction of displacement. The force transfers energy to the object, increasing its kinetic energy. Example: an engine pushing a car forward.
- Negative work (θ > 90°): the force has a component opposite to the displacement. The force removes energy from the object. Example: friction opposing a sliding box; gravity acting on a rising object.
- Zero work (θ = 90°): the force is perpendicular to the displacement. No energy is transferred. Example: normal force on a horizontally moving box; centripetal force on an orbiting satellite.
The sign of work tells you the direction of energy transfer: positive work adds energy to the object; negative work removes it.
Example — Positive and negative work
A 5.0 kg block slides 3.0 m down a frictionless ramp. Gravity does positive work because the force has a component along displacement. Normal force does zero work because it is perpendicular. If friction is present, it does negative work because it opposes displacement.
14.5 The Work Equation
When a constant force F acts on an object that undergoes displacement d, with angle θ between the force vector and displacement vector, the work done is:
W = Fd cosθ
The cosine factor extracts the component of force along the displacement. When θ = 0°, force and displacement are parallel and W = Fd. When θ = 90°, the force is perpendicular and W = 0. When θ = 180°, the force is antiparallel and W = -Fd.
| Angle θ | cosθ | Work | Physical Meaning |
|---|---|---|---|
| 0° | 1.0 | W = +Fd | Maximum positive; force fully in direction of motion |
| 30° | 0.866 | W = 0.866 Fd | Positive; component along motion |
| 60° | 0.500 | W = 0.500 Fd | Positive; smaller component |
| 90° | 0 | W = 0 | No work; perpendicular force |
| 120° | -0.500 | W = -0.500 Fd | Negative; opposing component |
| 180° | -1.0 | W = -Fd | Maximum negative; force fully opposes motion |
Worked Example 14.1 — Pulling a Crate at an Angle
A rope pulls a 20.0 kg crate 8.0 m across a floor. The rope makes 35° with the horizontal. Tension T = 90.0 N. How much work does the tension do?
- Identify: F = T = 90.0 N; d = 8.0 m; θ = 35°.
- Apply: W = Fd cosθ = 90.0 × 8.0 × cos 35° = 720 × 0.819 = 590 J.
- Check: Units are N × m = J; the work is positive because tension has a component along displacement.
Worked Example 14.2 — Work Done by Multiple Forces
The crate above also has friction fk = 30.0 N. Find the work done by friction, normal force, gravity, and the net work.
- Work by friction: Wf = 30.0 × 8.0 × cos 180° = -240 J.
- Work by normal: WN = N × 8.0 × cos 90° = 0 J.
- Work by gravity: Wg = mg × 8.0 × cos 90° = 0 J.
- Net work: Wnet = 590 + (-240) + 0 + 0 = 350 J.
14.6 Energy as the Capacity to Do Work
Energy is the capacity of a system to do work. A moving car has energy and can exert forces over a distance, such as crumpling metal in a crash. A raised weight has energy and can do work as it falls. A compressed spring has energy and can push objects when released.
Energy is conserved: it cannot be created or destroyed, only transferred between systems or converted between forms. This law — the law of conservation of energy — is one of the most fundamental laws in all of physics. Work is the mechanism by which energy is transferred: when a force does work on an object, it transfers energy to or from that object.
The SI unit of energy is the joule (J = N·m = kg·m²/s²). This is also the unit of work, because work measures the amount of energy transferred.
| Energy Form | Example | Rough Magnitude |
|---|---|---|
| Kinetic energy | 60 kg person running at 4 m/s | 480 J |
| Gravitational PE | Same person climbing 10 m stairs | 5886 J |
| Chemical energy | One food Calorie, kcal | 4186 J |
| Thermal energy | Heating 1 kg water 1°C | 4186 J |
| Nuclear energy | Fission of 1 kg U-235 | ~8 × 10¹³ J |
| Rest-mass energy | 1 kg of matter, E = mc² | 9 × 10¹⁶ J |
14.7 Kinetic Energy
Kinetic energy (KE) is the energy an object possesses due to its motion. It depends on both mass and velocity:
KE = ½mv²
Kinetic energy is a scalar — it has no direction. It depends on speed, the magnitude of velocity, not velocity itself. Two objects moving in opposite directions at the same speed have the same kinetic energy.
The velocity dependence is quadratic: doubling the speed quadruples the kinetic energy. This is why collision safety is so much more important at high speeds: a car at 100 km/h has four times the kinetic energy of the same car at 50 km/h, meaning it needs four times the stopping distance for the same braking force and absorbs four times the energy in a crash.
| Object | Mass | Speed | Kinetic Energy |
|---|---|---|---|
| Person walking | 70 kg | 1.4 m/s | 69 J |
| Person running | 70 kg | 4.0 m/s | 560 J |
| Car at 60 km/h | 1200 kg | 16.7 m/s | 167,280 J ≈ 167 kJ |
| Car at 120 km/h | 1200 kg | 33.3 m/s | 666,708 J ≈ 667 kJ |
| Bullet | 0.010 kg | 900 m/s | 4050 J |
| Passenger jet | 200,000 kg | 250 m/s | 6.25 × 10⁹ J |
Example — KE calculation
A 0.145 kg baseball pitched at 42.0 m/s has KE = ½ × 0.145 × (42.0)² = 0.0725 × 1764 = 127.9 J ≈ 128 J.
14.8 The Work-Energy Theorem
The work-energy theorem is one of the most powerful results in mechanics. It states: the net work done on an object equals the change in the object’s kinetic energy.
W_net = ΔKE = KE_f - KE_i = ½mv_f² - ½mv_i²
This theorem follows directly from Newton’s second law combined with kinematics: applying F = ma and the kinematic equation vf² = vi² + 2aΔx yields Wnet = FΔx = ½mvf² - ½mvi². The work-energy theorem is not a new law; it is a consequence of Newton’s second law expressed in energy language.
The theorem is useful because it bypasses the need to know the acceleration or the time — only the net work and the velocities matter. It is especially powerful for variable forces, curved paths, and problems where the force changes in magnitude or direction along the path.
Worked Example 14.3 — Using the Work-Energy Theorem to Find Final Speed
A 5.0 kg block starts from rest and a net force of 25.0 N pushes it 8.0 m. Find the final speed.
- Net work: Wnet = F × d × cos 0° = 25.0 × 8.0 = 200 J.
- Work-energy theorem: Wnet = ½mvf² - ½mvi² = ½mvf² - 0.
- Solve: 200 = ½ × 5.0 × vf², so vf² = 80 and vf = 8.94 m/s.
- Check with kinematics: a = F/m = 5.0 m/s²; vf² = 2(5.0)(8.0) = 80.
Worked Example 14.4 — Finding Stopping Distance
A 1200 kg car moving at 25.0 m/s brakes with a friction force of 8000 N. How far does it travel before stopping?
- Initial KE: KEi = ½ × 1200 × (25.0)² = 375,000 J.
- Final KE: KEf = 0, because the car stops.
- Work-energy theorem: Wnet = KEf - KEi = 0 - 375,000 = -375,000 J.
- Work by friction: Wf = -fkd = -375,000 J, so d = 375,000 / 8000 = 46.9 m.
14.9 Power and Rate of Doing Work
Power is the rate at which work is done, or equivalently, the rate at which energy is transferred. Two machines that do the same total work differ in power if one does it faster than the other.
P = W / t
An equivalent expression in terms of force and velocity: when a constant force F acts on an object moving at velocity v in the direction of the force:
P = Fv
| Power Source | Approximate Power |
|---|---|
| Human at rest, metabolic | 80 W |
| Human cycling hard | 200-400 W |
| Human sprinting, brief | ~1000 W |
| 1 horsepower, 1 hp | 746 W |
| Small car engine | 50-100 kW |
| Sports car engine | 200-500 kW |
| Large power plant | 1 GW, 10⁹ W |
| Sun’s total output | 3.8 × 10²⁶ W |
Example — Power calculation
A 60.0 kg person climbs a 10.0 m staircase in 20.0 s. Work against gravity = mgh = 60.0 × 9.81 × 10.0 = 5886 J. Power = W/t = 5886/20.0 = 294 W.
14.10 Mechanical Power in Real Systems
In real machines, not all input power appears as useful output power. Friction, air resistance, and other dissipative forces convert some mechanical energy to thermal energy. The efficiency of a machine quantifies this:
efficiency = W_out / W_in = P_out / P_in
No real machine is 100% efficient. Gasoline engines are typically 20-35% efficient; most energy becomes heat. Electric motors can be 85-95% efficient. The human body is about 25% efficient at converting food energy to mechanical work.
Example — Engine efficiency
A car engine receives fuel energy at 80.0 kW and delivers 28.0 kW to the wheels. Efficiency = 28.0/80.0 = 0.35 = 35%. The remaining 52.0 kW heats the engine, exhaust, and radiator.
The P = Fv form is particularly useful for vehicles. The maximum speed of a car occurs when the engine’s power equals the power dissipated by drag and friction: Pengine = Fdrag × vmax.
Example — Maximum speed from power
A car engine delivers P = 100 kW. At maximum speed, drag force = 800 N. vmax = P/F = 100,000/800 = 125 m/s = 450 km/h. This is an extreme example; real limits include aerodynamics and safety.
14.11 Energy Transfer and Transformation
Work is one mechanism for transferring energy between objects. When a force does positive work on an object, energy is transferred from the agent exerting the force to the object. When a force does negative work, energy flows the other direction.
Energy also transforms between forms without being transferred between objects. A falling ball converts gravitational potential energy to kinetic energy. A braking car converts kinetic energy to thermal energy. A battery converts chemical energy to electrical energy. In every transformation, the total energy is conserved; only its form changes.
| Process | Energy Input Form | Energy Output Form |
|---|---|---|
| Object speeding up, engine | Chemical, fuel | Kinetic |
| Object slowing down, brakes | Kinetic | Thermal, heat |
| Object rising | Kinetic or work | Gravitational PE |
| Object falling | Gravitational PE | Kinetic |
| Spring launching object | Elastic PE, spring | Kinetic |
| Light bulb | Electrical | Light + thermal |
| Photosynthesis | Light, solar | Chemical |
The work-energy theorem is the mechanics version of energy conservation: all energy transferred into an object as net work appears as kinetic energy. Chapter 15 extends this to include potential energy stored in gravitational and spring systems, leading to the full conservation of mechanical energy principle.
14.12 Common Mistakes in Work and Energy Problems
Mistake 1: Using total force instead of net work
The work-energy theorem uses the net work done by all forces. Calculate the work done by each force separately, then sum them with signs to get the net work. Do not use the total magnitude of all forces without regard to direction.
Mistake 2: Using the wrong angle in W = Fd cosθ
θ is the angle between the force vector and the displacement vector — not the angle between the force and the horizontal, or any other reference. Draw both vectors from a common tail and measure the angle between them directly.
Mistake 3: Forgetting that perpendicular forces do zero work
Normal force on a horizontally moving object, centripetal force on a circular orbit, and gravity on a horizontally moving object do zero work. These forces may change the direction of motion but do not transfer energy.
Mistake 4: Treating kinetic energy as a vector
KE = ½mv² is always positive or zero. It does not have direction. Do not assign a positive or negative sign to kinetic energy based on the direction of motion.
Mistake 5: Confusing work and power
Work is the total energy transferred, measured in joules. Power is the rate of transfer, measured in watts, or joules per second. A machine with high power does work quickly; it may or may not do more total work than a lower-power machine.
Mistake 6: Forgetting to convert units
km/h must be converted to m/s before using energy formulas. A car at 100 km/h = 27.8 m/s. KE = ½ × 1200 × (27.8)² = 463,000 J — not ½ × 1200 × (100)².
Chapter Summary
Work is the transfer of energy by a force; the work-energy theorem links the net work on an object to its change in kinetic energy. Power is the rate of work done.
- Work: W = Fd cosθ; units = joules (J = N·m). Positive when force has component along displacement; negative when opposing; zero when perpendicular.
- Three conditions for work: a force must act, the object must move, and the force must have a component along the displacement.
- Kinetic energy: KE = ½mv²; scalar; always ≥ 0; quadrupled when speed doubles.
- Work-energy theorem: Wnet = ΔKE = ½mvf² - ½mvi².
- Power: P = W/t; units = watts (W = J/s). Also P = Fv for a constant force along velocity.
- Efficiency = Pout/Pin; always ≤ 1; no real machine is 100% efficient.
- Energy is conserved; work is the mechanism of transfer; kinetic energy is one of its forms.
Key Terms
- work
- The energy transferred to or from an object by a force acting over a displacement: W = Fd cosθ; SI unit: joule (J).
- joule (J)
- The SI unit of work and energy: 1 J = 1 N·m = 1 kg·m²/s².
- kinetic energy (KE)
- Energy possessed by an object due to its motion: KE = ½mv²; scalar; always ≥ 0.
- work-energy theorem
- The net work done on an object equals its change in kinetic energy: Wnet = ΔKE.
- power
- The rate at which work is done or energy is transferred: P = W/t; SI unit: watt (W = J/s).
- watt (W)
- The SI unit of power: 1 W = 1 J/s = 1 kg·m²/s³.
- efficiency
- The ratio of useful output energy or power to total input energy or power; dimensionless; always ≤ 1.
- energy
- The capacity of a system to do work; conserved in all physical processes; SI unit: joule (J).
- positive work
- Work done when the force has a component in the direction of displacement, 0° ≤ θ < 90°; increases the object’s kinetic energy.
- negative work
- Work done when the force has a component opposing displacement, 90° < θ ≤ 180°; decreases the object’s kinetic energy.
Concept Review Questions
- Define work in the physics sense. Give two examples from everyday life where effort is expended but no physical work is done.
- What three conditions must be met for a force to do work on an object?
- A centripetal force acts on a satellite in circular orbit. Does it do work? Explain.
- In the equation W = Fd cosθ, what does the angle θ represent? What happens to work as θ increases from 0° to 180°?
- Explain in your own words what the work-energy theorem says. How does it differ from Newton’s second law?
- A car moving at 100 km/h has how many times the kinetic energy of the same car at 50 km/h? Explain why.
- What is the difference between work and power? Give an example where one person does more work but another has greater power.
- What does it mean for a machine to be 80% efficient? Where does the other 20% of energy go?
- Friction does negative work on a sliding box. What happens to that energy — is it destroyed?
- Can an object have kinetic energy and zero velocity simultaneously? Can it have zero kinetic energy and non-zero velocity?
Practice Problems
Work Problems
Show all work. Include the angle used in each calculation.
- A force of 50.0 N is applied horizontally to a 15.0 kg box, moving it 6.0 m. Find the work done by the force.
- A rope pulls a 10.0 kg sled 15.0 m at 25° above horizontal with tension 40.0 N. Find the work done by: (a) tension; (b) gravity; (c) normal force.
- A 70.0 kg person climbs 12.0 m vertically up a ladder. Find the work done by: (a) gravity on the person; (b) the person’s legs against gravity.
- A friction force of 25.0 N acts on a 5.0 kg block sliding 4.0 m. Find the work done by friction.
- Find the net work done on a 8.0 kg box pushed 5.0 m by a 30.0 N horizontal force with 12.0 N friction opposing.
- A 1200 kg car decelerates from 20.0 m/s to 8.0 m/s over 50.0 m. Find the net work done on the car.
Kinetic Energy Problems
- Find the kinetic energy of: (a) a 0.145 kg baseball at 40.0 m/s; (b) a 70.0 kg runner at 8.0 m/s; (c) a 1500 kg car at 30.0 m/s.
- A 5.0 kg object has 360 J of kinetic energy. What is its speed?
- A 1200 kg car accelerates from 0 to 27.8 m/s, or 100 km/h. Find the change in kinetic energy.
- Use the work-energy theorem to find the final speed of a 3.0 kg block starting from rest after 40.0 J of net work is done on it.
- A 0.50 kg ball rolls along a floor. Friction does -15.0 J of work on it. If it started at 6.0 m/s, what is its final speed?
- A car moving at 25.0 m/s brakes with a net force of 8000 N. Use the work-energy theorem to find how far it travels before stopping. Mass = 1200 kg.
Power Problems
- A motor does 12,000 J of work in 60.0 s. What is its power output?
- A 60.0 kg athlete runs up a 10.0 m staircase in 8.0 s. Find: (a) work done against gravity; (b) power output.
- An electric motor with power 2.5 kW runs for 3.0 minutes. How much work does it do?
- A car engine exerts a driving force of 3500 N at 28.0 m/s, constant speed. Find the engine’s power output.
- A 75.0 kg cyclist produces 250 W of power. At what speed does she travel against a 200 N drag force? Hint: P = Fv.
- Two machines both do 10,000 J of work. Machine A takes 20.0 s; Machine B takes 50.0 s. Find the power of each. Which would you prefer to have move a heavy object quickly?
Mixed Work-Energy Analysis
- A 2.0 kg block starts at rest and is pulled 4.0 m along a frictionless surface by a force of 10.0 N at 30° above horizontal. Find: (a) work done by the applied force; (b) final KE; (c) final speed.
- A 1500 kg car traveling at 20.0 m/s brakes to a stop. (a) How much kinetic energy is lost? (b) If braking takes 40.0 m, what average friction force acted? (c) What is the power dissipated if braking takes 4.0 s?
- A 0.20 kg ball is thrown upward at 15.0 m/s. Using the work-energy theorem, with gravity doing work = -mgh: (a) how high does the ball rise? (b) What is its speed at 8.0 m above the launch point?
- An engine rated at 80.0 kW powers a 2000 kg vehicle against a 600 N drag. (a) What is the maximum speed? (b) If the engine is 30% efficient, what is the fuel power input?
- Challenge: A block, m = 4.0 kg, starts from rest at the top of a 35°, 6.0 m ramp. μk = 0.20. Using the work-energy theorem, find work by gravity, normal, and friction separately, then find the speed at the bottom.
Chapter 14 Checkpoint
| I can… | Section |
|---|---|
| State the three conditions required for a force to do work | 14.3 |
| Calculate work using W = Fd cosθ and identify θ correctly | 14.5 |
| Determine whether work is positive, negative, or zero from the angle | 14.4 |
| Calculate kinetic energy using KE = ½mv² | 14.7 |
| Apply the work-energy theorem to find speed or distance | 14.8 |
| Calculate power using P = W/t and P = Fv | 14.9 |
| Calculate mechanical efficiency | 14.10 |
| Identify common energy transformation chains | 14.11 |
| Avoid the six most common work and energy mistakes | 14.12 |