solve problems using P = W / t

Energy Conservation & Power (Cambridge IGCSE/A‑Level 9702)

1. Syllabus‑Coverage Checklist

2. Prerequisite Recap (Topics 1‑5 of the syllabus)

• Force & displacement: Work is the scalar product of the force vector \(\mathbf{F}\) and the displacement vector \(\mathbf{s}\):

  \[

  W = \mathbf{F}\!\cdot\!\mathbf{s}=Fs\cos\theta\quad(\text{J})

  \]

  \(\theta\) is the angle between \(\mathbf{F}\) and \(\mathbf{s}\).

• Kinematics: Constant‑acceleration equations give the distance travelled under a constant net force. From these we obtain the gravitational‑potential‑energy form

  \[

  E_{p}=mgh

  \]

  (derived from the work done against the weight \(mg\) over a height \(h\)).

• Newton’s 2nd law: \(F = ma\) allows replacement of force by mass × acceleration when the motion is known.

3. Key Definitions

4. Work – Energy – Power (Core AS Content)

4.1 Work‑Energy Theorem

For a particle acted on by a net force \(\Sigma\mathbf{F}\) over a displacement \(\mathbf{s}\):

\[

W{\text{net}} = \Delta E{k} = \tfrac12 m v{f}^{2} - \tfrac12 m v{i}^{2}

\]

If non‑conservative forces (e.g. friction) are present, the theorem is written as

\[

W{\text{conservative}} + W{\text{non‑conservative}} = \Delta E_{k}

\]

where \(W_{\text{non‑conservative}}\) is often negative (energy lost as heat).

4.2 Sign of Work

• Positive work: force component has a component in the direction of motion (e.g. pushing a sled forward).
• Negative work**: force opposes the motion (e.g. kinetic‑friction force, braking). Negative work removes kinetic energy from the system.

Example (negative work): A 1500 kg car decelerates from 20 m s⁻¹ to rest under a constant braking force.

\[

W = \Delta E_{k} = \tfrac12 (1500)(0^{2} - 20^{2}) = -3.0\times10^{5}\ \text{J}

\]

The braking force does \(-3.0\times10^{5}\) J of work.

4.3 Power from Work

\[

P = \frac{W}{t}\qquad\text{or}\qquad P = \mathbf{F}\!\cdot\!\mathbf{v}

\]

The second form shows that power is the scalar product of force and instantaneous velocity.

4.4 Efficiency Factor

\[

P{\text{useful}} = \eta\,P{\text{input}},\qquad

W{\text{useful}} = \eta\,W{\text{input}}

\]

Typical values: \(\eta{\text{electric motor}}\approx0.8\), \(\eta{\text{human}}\approx0.25\).

4.5 Step‑by‑Step Procedure for \(P = \dfrac{W}{t}\)

1. Identify the unknown: power, work (energy), or time.
2. List all known quantities with units.
3. Convert every quantity to SI units (e.g. kJ → J, min → s).
4. Choose the appropriate rearranged formula:

   • Find work: \(W = P\,t\)
   • Find time: \(t = \dfrac{W}{P}\)
   • Find power: \(P = \dfrac{W}{t}\)

5. Insert numbers, keep track of significant figures, and check the sign of work.
6. Validate the answer with an energy‑conservation check or by estimating the order of magnitude.

4.6 Worked Examples

Example 1 – Thermal Power (Heater)

Question: A 1500 W electric heater raises 2.0 kg of water from 20 °C to 80 °C. How long does it take? (\(c_{\text{water}} = 4180\ \text{J kg}^{-1}\text{K}^{-1}\)).

1. Energy required:

   \[

   W = mc\Delta T = (2.0)(4180)(80-20)=5.02\times10^{5}\ \text{J}

2. Time:

   \[

   t = \frac{W}{P}= \frac{5.02\times10^{5}}{1500}=3.34\times10^{2}\ \text{s}\approx5.6\ \text{min}

   \]

Example 2 – Mechanical Power (Cyclist)

Question: An 80 kg cyclist climbs a 300 m hill at a constant speed of 3 m s⁻¹. Find the average power output, neglecting losses.

1. Work against gravity:

   \[

   W = mgh = (80)(9.81)(300)=2.36\times10^{5}\ \text{J}

2. Time taken:

   \[

   t = \frac{\text{distance}}{\text{speed}} = \frac{300}{3}=100\ \text{s}

   \]

3. Power:

   \[

   P = \frac{W}{t}= \frac{2.36\times10^{5}}{100}=2.36\times10^{3}\ \text{W}

   \]

Example 3 – Negative Work (Friction)

A 10 kg block slides 5 m across a rough horizontal surface. The coefficient of kinetic friction is \(\mu_k = 0.30\). Find the work done by friction and the average power if the motion takes 4 s.

\[

F{\text{fr}} = \muk mg = 0.30(10)(9.81)=29.4\ \text{N}

\]

\[

W{\text{fr}} = -F{\text{fr}}s = -(29.4)(5) = -1.47\times10^{2}\ \text{J}

\]

\[

P = \frac{W}{t}= \frac{-1.47\times10^{2}}{4}= -3.7\times10^{1}\ \text{W}

\]

(The negative sign indicates that friction removes energy from the block.)

5. Electrical Power (DC Circuits – AS Topics 9 & 10)

5.1 Fundamental Power Relations

\[

P = VI = I^{2}R = \frac{V^{2}}{R}

\]

5.2 Kirchhoff’s Laws (required for syllabus)

• Current law (KCL): The algebraic sum of currents at a junction is zero.
• Voltage law (KVL): The algebraic sum of potential differences around any closed loop is zero.

5.3 Internal Resistance of a Cell

For a source with emf \(\mathcal{E}\) and internal resistance \(r\) delivering current \(I\) to an external resistance \(R\):

\[

V = \mathcal{E} - Ir,\qquad

P_{\text{delivered}} = VI = I^{2}R

\]

The power dissipated inside the cell is \(P_{\text{int}} = I^{2}r\). Efficiency:

\[

\eta = \frac{I^{2}R}{I^{2}(R+r)} = \frac{R}{R+r}

\]

5.4 Worked Electrical Example

Question: A 12 V battery has emf \(\mathcal{E}=12.5\ \text{V}\) and internal resistance \(r=0.5\ \Omega\). It powers a lamp of resistance \(R=6\ \Omega\). Find the terminal voltage, the current, the power delivered to the lamp and the efficiency.

\[

I = \frac{\mathcal{E}}{R+r}= \frac{12.5}{6.5}=1.92\ \text{A}

\]

\[

V_{\text{terminal}} = \mathcal{E} - Ir = 12.5 - (1.92)(0.5)=11.5\ \text{V}

\]

\[

P_{\text{lamp}} = I^{2}R = (1.92)^{2}(6)=22.1\ \text{W}

\]

\[

\eta = \frac{P_{\text{lamp}}}{\mathcal{E}I}= \frac{22.1}{12.5\times1.92}=0.92\;(92\%)

\]

6. Waves – Superposition (AS Topic 7)

6.1 Intensity and Power

For a wave travelling with speed \(v\) and area \(A_{\text{wave}}\) of the wave front:

\[

I = \frac{P}{A{\text{wave}}}\quad\Longrightarrow\quad P = I\,A{\text{wave}}

\]

Intensity is proportional to the square of the amplitude:

\[

I \propto A^{2}

\]

6.2 Stationary (Standing) Waves

• Formed by the superposition of two waves of equal amplitude travelling in opposite directions.
• Nodes: points of zero displacement; Antinodes: points of maximum displacement.
• For a string of length \(L\) fixed at both ends, the allowed wavelengths are \(\lambda_n = \dfrac{2L}{n}\) ( \(n=1,2,3,\dots\) ).

6.3 Interference & Power

When two coherent sources of equal amplitude \(A\) interfere, the resultant intensity at a point is

\[

I = I{1}+I{2}+2\sqrt{I{1}I{2}}\cos\phi

\]

where \(\phi\) is the phase difference. Constructive interference (\(\phi=0\)) gives \(I=4I_{1}\); destructive (\(\phi=\pi\)) gives \(I=0\).

6.4 Example – Sound Intensity

A speaker produces a sound intensity of \(1.0\times10^{-3}\ \text{W m}^{-2}\) at a distance of 2 m. What is the total acoustic power output?

\[

A_{\text{sphere}} = 4\pi r^{2}=4\pi(2)^{2}=50.3\ \text{m}^{2}

\]

\[

P = I A = (1.0\times10^{-3})(50.3)=5.0\times10^{-2}\ \text{W}

\]

7. Deformation of Solids (AS Topic 6)

7.1 Stress, Strain & Young’s Modulus

\[

\text{Stress } (\sigma) = \frac{F}{A}\qquad

\text{Strain } (\varepsilon) = \frac{\Delta L}{L_{0}}

\]

\[

\text{Young’s modulus } (Y) = \frac{\sigma}{\varepsilon}

\]

Units: \(\sigma\) in pascals (Pa), \(\varepsilon\) dimensionless, \(Y\) in Pa.

7.2 Elastic‑Potential Energy of a Spring

\[

E_{e}= \tfrac12 kx^{2}

\]

where \(k\) is the spring constant (N m⁻¹) and \(x\) the extension/compression.

7.3 Experimental Determination of Young’s Modulus

Typical procedure (AO3 focus):

1. Measure the original length \(L_{0}\) and cross‑sectional area \(A\) of a metal rod.
2. Apply a series of known forces \(F\) (using masses) and record the corresponding extensions \(\Delta L\).
3. Plot \(\sigma = F/A\) against \(\varepsilon = \Delta L/L_{0}\). The gradient of the linear region gives \(Y\).
4. Analyse uncertainties (reading error, mass tolerance) and calculate the percentage error of the experimental \(Y\) compared with the accepted value.

7.4 Example – Spring Constant

A spring is stretched 0.12 m by a force of 24 N. Find \(k\) and the elastic‑potential energy stored.

\[

k = \frac{F}{x}= \frac{24}{0.12}=200\ \text{N m}^{-1}

\]

\[

E_{e}= \tfrac12 kx^{2}= \tfrac12 (200)(0.12)^{2}=1.44\ \text{J}

\]

8. Common Pitfalls

• Forgetting to convert minutes or hours to seconds before using \(P=W/t\).
• Omitting the \(\cos\theta\) factor in \(W=Fs\cos\theta\); this leads to sign errors.
• Assuming all work is done by conservative forces – always check for friction or air resistance.
• Using the wrong form of electrical power (e.g., forgetting internal resistance when the circuit is not ideal).
• Neglecting the efficiency factor when the question states that the device is not 100 % efficient.
• Mixing up intensity (W m⁻²) with power (W) – remember to multiply by the appropriate area.

9. Practical Skills (AO3 – Paper 3/5)

Students should be able to:

1. Plan*: Sketch a clear circuit diagram, label emf, internal resistance, and load; decide which quantities to measure (V, I, t) and the method of measurement (voltmeter, ammeter, stop‑watch).
2. Collect*: Record multiple readings, use appropriate ranges, and note environmental conditions (temperature, friction surface).
3. Analyse*: Convert raw data to SI units, calculate work, power and efficiency, propagate uncertainties (e.g., \(\Delta P/P = \sqrt{(\Delta V/V)^{2}+(\Delta I/I)^{2}}\)).
4. Evaluate*: Discuss sources of error (instrumental, systematic, assumptions such as neglecting friction), suggest improvements, and comment on how the uncertainties affect the final result.

10. Linking Power to Energy Conservation

In a steady‑state system the total energy does not change, so the net power flow is zero:

\[

\sum P{\text{in}} - \sum P{\text{out}} = \frac{dE_{\text{total}}}{dt}=0

\]

Consequently, the sum of all power inputs equals the sum of all power outputs. This principle underlies generator‑engine balances, thermal‑engine efficiency calculations, and fluid‑power problems.

11. A‑Level Extensions (Syllabus Items 12‑25)

• Circular motion: Uniform circular motion → centripetal force ⟂ displacement → power = 0.
• Gravitational fields: Potential energy \(U = -\dfrac{GMm}{r}\); power of a satellite = \(\dfrac{dU}{dt}\).
• Thermodynamics: Heat‑transfer rate \(\dot Q = \dfrac{dQ}{dt}\); first law in power form \(\dot Q{\text{in}}-\dot Q{\text{out}} = \dfrac{dU}{dt}+P_{\text{work}}\).
• Fluids: Power = pressure × volume‑flow rate, \(P = p\,Q\) (e.g., pump).
• Oscillations: Mechanical power in a simple harmonic oscillator \(P = -kxv\); average power over a cycle is zero.
• Electric & magnetic fields: Energy density \(u = \tfrac12 \varepsilon{0}E^{2} + \tfrac12 \dfrac{B^{2}}{\mu{0}}\); power transferred to a magnetic field \(P = \mathbf{B}\cdot\frac{d\mathbf{M}}{dt}\).
• AC circuits: Instantaneous power \(p(t)=v(t)i(t)\); average power \(P_{\text{av}} = VI\cos\phi\) (power factor).
• Quantum & nuclear physics: Decay power \(P = \lambda N E_{\gamma}\); binding‑energy calculations.
• Medical & astronomical applications: Power output of a laser, luminosity of a star (\(L = 4\pi R^{2}\sigma T^{4}\)).

12. Quick Reference Table

13. Suggested Diagram