Energy conservation

1. Overview

The principle of energy conservation is the cornerstone of classical mechanics. It dictates that in a closed system, the total quantity of energy remains invariant over time, regardless of the internal transformations occurring. Work done serves as the precise mathematical link between dynamics (forces) and energetics (energy states). It represents the process by which a force transfers energy between objects or transforms it from one variety to another. By mastering the relationship between work, power, and efficiency, you can solve complex mechanical problems—such as those involving variable paths or resistive forces—where the direct application of Newton’s Second Law might be mathematically cumbersome.

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Key Definitions

• Work Done: The product of the force and the displacement in the direction of the force. It is a scalar quantity. (Unit: Joule, J).
• The Joule: The work done when a force of one newton moves its point of application a distance of one metre in the direction of the force. ($1\text{ J} = 1\text{ N m}$).
• Principle of Conservation of Energy: Energy cannot be created or destroyed; it can only be transformed from one form to another or transferred from one body to another. The total energy of a closed system remains constant.
• Efficiency: The ratio of the useful energy output from a system to the total energy input. It is a dimensionless number often expressed as a percentage.
• Power: The rate at which work is done or the rate at which energy is transferred. (Unit: Watt, W).
• The Watt: A rate of energy transfer of one joule per second. ($1\text{ W} = 1\text{ J s}^{-1}$).

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Content

3.1 Work Done ($W$)

Work is only performed when the point of application of a force moves through a displacement. Crucially, only the component of the force that acts parallel to the displacement vector contributes to the work done.

The General Equation: $$W = Fs \cos \theta$$

Where:

• $W$ = Work done (J)
• $F$ = Magnitude of the constant force (N)
• $s$ = Magnitude of the displacement (m)
• $\theta$ = The angle between the force vector F and the displacement vector s.

Vector Analysis of Work:

1. Force in the direction of motion ($\theta = 0^\circ$): $\cos 0^\circ = 1$. The work done is simply $W = Fs$. This results in a gain in energy for the object (e.g., accelerating a car).
2. Force perpendicular to motion ($\theta = 90^\circ$): $\cos 90^\circ = 0$. No work is done. Examples include the centripetal force in circular motion or the normal contact force on an object moving horizontally.
3. Force opposing motion ($\theta = 180^\circ$): $\cos 180^\circ = -1$. The work done is $W = -Fs$. This represents energy being removed from the object, usually dissipated as heat (e.g., work done by friction).

3.2 The Principle of Conservation of Energy

In any physical process, the total energy of an isolated system is conserved. While energy may change form (e.g., from gravitational potential energy to kinetic energy), the sum of all forms remains constant.

Energy Balance Equation: $$\text{Total Energy Input} = \text{Useful Energy Output} + \text{Wasted Energy Output}$$

In A-Level mechanics, we frequently analyze the exchange between Kinetic Energy ($E_k$) and Potential Energy ($E_p$). In the absence of resistive forces (like air resistance or friction): $$\Delta E_k + \Delta E_p = 0$$ This implies that any loss in $E_p$ results in an equivalent gain in $E_k$.

Dissipative Forces: In real-world systems, work is often done against resistive forces. This "wasted" work is typically converted into internal energy (thermal energy), which increases the temperature of the object and its surroundings. $$\text{Initial } E_k + \text{Initial } E_p = \text{Final } E_k + \text{Final } E_p + \text{Work Done against Friction}$$

3.3 Efficiency ($\eta$)

Efficiency is a measure of how effectively a device transforms input energy into a desired useful output. Due to the Second Law of Thermodynamics, no macroscopic machine can be 100% efficient; some energy is always dissipated as heat.

Mathematical Expressions: $$\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}}$$ $$\text{Efficiency} = \frac{\text{Useful power output}}{\text{Total power input}}$$

Note: Efficiency is a ratio. To express it as a percentage, multiply the result by 100.

3.4 Power ($P$)

Power quantifies the "speed" of energy transfer. Two motors might lift the same weight to the same height (doing the same work), but the motor that does it faster is more powerful.

The Fundamental Equation: $$P = \frac{W}{t}$$

Where:

• $P$ = Power (W)
• $W$ = Work done (J)
• $t$ = Time taken (s)

3.5 Derivation of $P = Fv$

The Cambridge syllabus requires you to derive the relationship between power, force, and velocity for an object moving at a constant speed.

1. Definition of Power: Start with the rate of work done: $$P = \frac{W}{t}$$
2. Substitution of Work: Substitute the formula for work done ($W = Fs$), assuming the force is applied in the direction of displacement: $$P = \frac{Fs}{t}$$
3. Introduction of Velocity: Recognize that for an object moving at a constant speed, velocity is the rate of change of displacement ($v = \frac{s}{t}$): $$P = F \left( \frac{s}{t} \right)$$ $$P = Fv$$

Application Note: In problems involving vehicles at constant velocity, $F$ represents the motive force (the forward force provided by the engine). Because the velocity is constant, the motive force must be equal in magnitude to the total resistive forces (drag and friction) acting on the vehicle.

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Worked Example 1 — Efficiency and Work on an Incline

A crate of mass $85.0\text{ kg}$ is pulled at a constant speed up a ramp inclined at $25.0^\circ$ to the horizontal. The length of the ramp is $15.0\text{ m}$. A winch provides a constant pulling force of $420\text{ N}$ parallel to the ramp to move the crate from the bottom to the top.

Calculate: (a) The total work done by the winch. (b) The useful work done (the gain in gravitational potential energy). (c) The efficiency of the system.

Part (a) Strategy: Work done is the force multiplied by the distance moved in the direction of the force. Here, the force and displacement are both parallel to the ramp. $$W_{total} = F \times s$$ $$W_{total} = 420\text{ N} \times 15.0\text{ m}$$ $$W_{total} = 6300\text{ J}$$ Answer: $6.30 \times 10^3\text{ J}$

Part (b) Strategy: The useful work is the energy required to lift the mass to the vertical height $h$. First, find $h$ using trigonometry: $h = 15.0 \sin(25.0^\circ) = 6.339\text{ m}$. $$W_{useful} = \Delta E_p = mgh$$ $$W_{useful} = 85.0 \times 9.81 \times 6.339$$ $$W_{useful} = 5285.8\text{ J}$$ Answer: $5.29 \times 10^3\text{ J}$

Part (c) Strategy: Efficiency is the ratio of useful output to total input. $$\text{Efficiency} = \frac{W_{useful}}{W_{total}}$$ $$\text{Efficiency} = \frac{5285.8}{6300} = 0.8390...$$ Answer: $83.9%$ (or $0.839$)

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Worked Example 2 — Power and Resistive Forces

A car of mass $1200\text{ kg}$ travels at a constant speed of $32.0\text{ m s}^{-1}$ along a level road. The engine develops a useful power of $45.0\text{ kW}$. The car then reaches a hill inclined at $4.00^\circ$ to the horizontal.

(a) Calculate the total resistive force acting on the car on the level road. (b) Assuming the resistive force remains the same, calculate the power the engine must now provide to maintain the same speed of $32.0\text{ m s}^{-1}$ up the hill.

Part (a) Strategy: At constant speed, the motive force $F$ equals the resistive force $R$. Use $P = Fv$. $$P = 45.0 \times 10^3\text{ W}$$ $$F = \frac{P}{v} = \frac{45000}{32.0}$$ $$F = 1406.25\text{ N}$$ Answer: $1410\text{ N}$ (3 s.f.)

Part (b) Strategy: To move up the hill at constant speed, the engine must overcome both the resistive force ($R$) and the component of the car's weight acting down the slope ($mg \sin \theta$). $$\text{Total Force Required } (F_{new}) = R + mg \sin \theta$$ $$F_{new} = 1406.25 + (1200 \times 9.81 \times \sin 4.00^\circ)$$ $$F_{new} = 1406.25 + 821.28 = 2227.53\text{ N}$$ Now calculate the new power: $$P_{new} = F_{new} \times v$$ $$P_{new} = 2227.53 \times 32.0 = 71280.96\text{ W}$$ Answer: $71.3\text{ kW}$

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Key Equations

Quantity	Equation	Symbols & Units	Status
Work Done	$W = Fs \cos \theta$	$F$(N), $s$(m), $\theta$(rad/deg)	Memorise
Power	$P = \frac{W}{t}$	$W$(J), $t$(s), $P$(W)	Data Sheet
Power (Motion)	$P = Fv$	$F$(N), $v$(m s⁻¹)	Derive/Memorise
Efficiency	$\eta = \frac{\text{Useful Output}}{\text{Total Input}}$	Ratio (no units)	Memorise
Potential Energy	$\Delta E_p = mgh$	$m$(kg), $g$($9.81\text{ m s}^{-2}$), $h$(m)	Data Sheet
Kinetic Energy	$E_k = \frac{1}{2}mv^2$	$m$(kg), $v$(m s⁻¹)	Data Sheet

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Common Mistakes to Avoid

• ❌ Wrong: Using $W = Fs$ where $s$ is the total distance traveled, even if the force is not in that direction.
  • ✅ Right: Always ensure $s$ is the displacement in the direction of the force, or use the $\cos \theta$ component.
• ❌ Wrong: Confusing the "motive force" with the "net force" in $P = Fv$.
  • ✅ Right: In $P = Fv$, $F$ is the specific force whose power you are calculating (usually the engine's driving force). If the object is at constant velocity, the net force is zero, but the power developed by the engine is not zero.
• ❌ Wrong: Forgetting to convert power from kW or MW to Watts before calculation.
  • ✅ Right: Standardize all units to SI base units ($1\text{ kW} = 10^3\text{ W}$, $1\text{ MW} = 10^6\text{ W}$) at the start of the problem.
• ❌ Wrong: Using $g = 10\text{ m s}^{-2}$.
  • ✅ Right: Always use $g = 9.81\text{ m s}^{-2}$ as specified in the Cambridge Data Booklet.
• ❌ Wrong: Calculating efficiency as a value greater than 1.
  • ✅ Right: Efficiency is always $\leq 1$ (or $100%$). If your calculation gives a higher value, you have likely swapped the input and output values.

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Exam Tips

1. The "Constant Speed" Keyword: When an exam question mentions "constant speed" or "terminal velocity," immediately recognize that the resultant force is zero. This means the forward (motive) force is exactly equal to the sum of all resistive forces.
2. Work Done by a Gas: Although covered in later topics, remember that work done can also be calculated from pressure-volume changes ($W = P \Delta V$). In mechanics, however, stick to $W = Fs \cos \theta$.
3. Resolving Weight on Slopes: This is a frequent source of errors.
   • Component of weight down the slope: $mg \sin \theta$.
   • Component of weight perpendicular to the slope: $mg \cos \theta$.
   • When calculating the work done against gravity, you can either use $W = (mg \sin \theta) \times \text{slope length}$ or simply $W = mg \times \text{vertical height}$. Both yield the same result.
4. Defining the Joule/Watt: If asked to define the Joule or Watt, do not just give the formula. Use the "One Newton, One Metre" or "One Joule per Second" phrasing to ensure you hit the mark scheme's specific requirements.
5. Significant Figures: Look at the data provided. If the mass is $5.0\text{ kg}$ (2 s.f.) and the height is $2.50\text{ m}$ (3 s.f.), your final answer should generally be given to 2 s.f. However, keep full precision in your calculator during intermediate steps to avoid rounding errors.