Power – Work Done per Unit Time

Objective

By the end of this topic students will be able to:

• State the definitions and SI units of work, energy and power (scalar or vector nature where appropriate).
• Derive the kinetic‑energy, gravitational‑potential‑energy and elastic‑potential‑energy expressions from the definition of work.
• Apply the work‑energy theorem and the various power formulae to mechanical and electrical situations.
• Calculate the efficiency of real devices and interpret power‑rating information.
• Plan and execute practical measurements of power (Paper 3 skills), including uncertainty analysis.

Key Concepts

1. Work (W)

• Work is the **scalar** product of a force and the displacement through which it acts.
• Constant force: \(W = F\,s\cos\theta = \vec F\!\cdot\!\vec s\)
• Variable force: \(W = \displaystyle\int\vec F\!\cdot\!d\vec s\)
• SI unit: joule (J) \(1\;\text{J}=1\;\text{N·m}\)

2. Energy (E)

• Energy is the capacity to do work.
• Kinetic energy (derived from work to accelerate a mass): \(K = \dfrac12 mv^{2}\)
• Gravitational potential energy (work against weight): \(U_g = mgh\) (from \(W = \int mg\,dh\))
• Elastic (spring) potential energy (A‑Level): \(U_s = \dfrac12 kx^{2}\)
• SI unit: joule (J).

3. Energy‑Conservation Principle (Syllabus wording)

For an isolated system the total energy remains constant:

\[ \Delta E_{\text{total}} = 0 \]

In differential form this becomes a **power balance**:

\[ \frac{dE_{\text{total}}}{dt}=0\;\Longrightarrow\;\sum P_{\text{in}} = \sum P_{\text{out}} \]

4. Work‑Energy Theorem

The net work done on a particle equals the change in its kinetic energy:

\[ W_{\text{net}} = \Delta K = K_{f}-K_{i} \]

5. Power (P)

• Average (constant) power: \(P_{\text{avg}} = \dfrac{W}{\Delta t}= \dfrac{\Delta E}{\Delta t}\)
• Instantaneous power: \(P = \dfrac{dW}{dt}= \vec F\!\cdot\!\vec v\)
• Rotational power (A‑Level): \(P = \tau\,\omega\) (torque × angular speed)
• SI unit: watt (W) \(1\;\text{W}=1\;\text{J·s}^{-1}\)

6. Electrical Power

Direct‑Current (DC)

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

Alternating‑Current (AC)

• RMS values: \(V_{\text{rms}} = \dfrac{V_{\text{peak}}}{\sqrt{2}}\), \(I_{\text{rms}} = \dfrac{I_{\text{peak}}}{\sqrt{2}}\)
• Apparent power: \(S = V_{\text{rms}}I_{\text{rms}}\) (unit: VA)
• Real (average) power: \(P = V_{\text{rms}}I_{\text{rms}}\cos\phi\) (\(\cos\phi\) = power factor)
• Reactive power: \(Q = V_{\text{rms}}I_{\text{rms}}\sin\phi\) (unit: VAR)

7. Efficiency

Efficiency expresses how well a device converts input power into useful output power:

\[ \eta = \frac{P_{\text{out}}}{P_{\text{in}}}\times100\% \]

• Mechanical efficiency (incl. friction, air resistance, etc.)
• Electrical efficiency (e.g., power factor for AC devices)

Units & Conversions

Quantity	SI Unit	Symbol	Common multiples
Work / Energy	joule	J	1 kJ = 10³ J
Power	watt	W	1 kW = 10³ W, 1 MW = 10⁶ W
Force	newton	N	1 kN = 10³ N
Voltage	volt	V	1 kV = 10³ V
Current	ampere	A	1 mA = 10⁻³ A
Resistance	ohm	Ω	–

Practical Measurement of Power (Paper 3 Skills)

1. Connect a calibrated watt‑meter (or a voltmeter and ammeter) to the circuit and read \(V\) and \(I\) simultaneously.
2. For DC circuits compute \(P = VI\). Record the uncertainties in \(V\) and \(I\); propagate them to obtain \(\Delta P\) using \[ \frac{\Delta P}{P}= \sqrt{\left(\frac{\Delta V}{V}\right)^{2}+\left(\frac{\Delta I}{I}\right)^{2}} . \]
3. For AC circuits measure rms values and the phase angle \(\phi\) (phase‑angle meter or oscilloscope). Real power is \(P = V_{\text{rms}}I_{\text{rms}}\cos\phi\).
4. Determine efficiency by measuring both the input power (from the source) and the useful output power (e.g., mechanical work on a dynamometer or lift height).
5. Report results with correct units, appropriate significant figures and a discussion of systematic errors (internal resistance of meters, contact resistance, etc.).

Worked Examples

Example 1 – Mechanical Power (AS)

Problem: A 1500 kg car accelerates from rest to \(20\;\text{m s}^{-1}\) in 10 s. Find the average power delivered by the engine.

1. \(\displaystyle \Delta K = \frac12 mv^{2}= \frac12(1500)(20)^{2}=3.0\times10^{5}\;\text{J}\)
2. \(\displaystyle P_{\text{avg}}=\frac{\Delta K}{\Delta t}= \frac{3.0\times10^{5}}{10}=3.0\times10^{4}\;\text{W}=30\;\text{kW}\)

Example 2 – Gravitational Power (A‑Level)

Problem: A 50 kg crate is lifted vertically at a constant speed of 2 m s⁻¹. Calculate the required power (ignore friction).

• Force needed: \(F = mg = 50\times9.8 = 490\;\text{N}\)
• Power: \(P = Fv = 490\times2 = 980\;\text{W}\)

Example 3 – Elastic Potential Energy & Rotational Power (A‑Level)

Problem: A light spring (constant \(k = 800\;\text{N m}^{-1}\)) is compressed 0.15 m and then released, driving a flywheel of moment of inertia \(I = 0.25\;\text{kg m}^{2}\) from rest to an angular speed of \(\omega = 30\;\text{rad s}^{-1}\). Show that the energy stored in the spring is sufficient and find the average power if the process takes 0.5 s.

• Elastic energy: \(U_s = \frac12 kx^{2}= \frac12(800)(0.15)^{2}=9.0\;\text{J}\)
• Rotational kinetic energy of flywheel: \(K_{\text{rot}} = \frac12 I\omega^{2}= \frac12(0.25)(30)^{2}=112.5\;\text{J}\)
• Because \(U_s < K_{\text{rot}}\) the spring alone cannot reach the final speed; an additional energy source would be required. (Illustrates the importance of checking energy budgets.)
• If the spring *could* supply the 112.5 J in 0.5 s, the average power would be \(P_{\text{avg}} = \dfrac{112.5}{0.5}=225\;\text{W}\).

Example 4 – DC Electrical Power (AS)

Problem: A heater is connected to a 240 V supply and draws 5 A. Find the electrical power consumed and the energy used after 3 h.

• Power: \(P = VI = 240\times5 = 1200\;\text{W}=1.2\;\text{kW}\)
• Energy: \(E = Pt = 1.2\;\text{kW}\times3\;\text{h}=3.6\;\text{kWh}=3.6\times3.6\times10^{6}\;\text{J}\)

Example 5 – AC Motor Power & Efficiency (A‑Level)

Problem: An AC motor is supplied with \(V_{\text{rms}}=230\;\text{V}\) and draws \(I_{\text{rms}}=8\;\text{A}\) with a power factor \(\cos\phi =0.85\). Determine the real power input and the motor’s efficiency if it delivers a mechanical output of 1.4 kW.

• Real power input: \(P_{\text{in}} = V_{\text{rms}}I_{\text{rms}}\cos\phi = 230\times8\times0.85 = 1.57\;\text{kW}\)
• Efficiency: \(\displaystyle \eta = \frac{1.4}{1.57}\times100\% \approx 89\%\)

Example 6 – Mechanical Efficiency (AO2)

Problem: A 20 kg block is pulled up a 5 m long incline that makes an angle of \(30^{\circ}\) with the horizontal. The coefficient of kinetic friction between block and plane is \(\mu_k = 0.20\). The block is moved at constant speed. Determine the mechanical efficiency of the system.

1. Weight component down the plane: \(W_{\parallel}=mg\sin\theta = 20\times9.8\times\sin30^{\circ}=98\;\text{N}\).
2. Normal reaction: \(N = mg\cos\theta = 20\times9.8\times\cos30^{\circ}=169.7\;\text{N}\).
3. Friction force: \(F_f = \mu_k N = 0.20\times169.7 = 33.9\;\text{N}\).
4. Total resisting force (gravity + friction): \(F_{\text{res}} = W_{\parallel}+F_f = 98+33.9 = 131.9\;\text{N}\).
5. Since the block moves at constant speed, the pulling force \(F_{\text{pull}} = F_{\text{res}}\). The **input work** over the distance \(s=5\;\text{m}\) is \(W_{\text{in}} = F_{\text{pull}}s = 131.9\times5 = 660\;\text{J}\).
6. The **useful work** is the increase in gravitational potential energy: \(W_{\text{out}} = mgh = mg(s\sin\theta)=20\times9.8\times(5\sin30^{\circ}) = 490\;\text{J}\).
7. Efficiency: \[ \eta = \frac{W_{\text{out}}}{W_{\text{in}}}\times100\% = \frac{490}{660}\times100\% \approx 74\%. \]

Common Misconceptions

• Energy vs. Power: Energy (J) is a quantity; power (W) is the *rate* at which energy is transferred.
• Average vs. Instantaneous: \(P = \dfrac{W}{t}\) gives only the average power. Instantaneous power requires the derivative or the dot‑product form \(P = \vec F\!\cdot\!\vec v\).
• Direction matters: In \(P = \vec F\!\cdot\!\vec v\) a force opposite to the motion gives negative power (energy is being removed from the system).
• Power factor is not a loss: A low \(\cos\phi\) means a large apparent power for the same real power; reactive power does not do useful work but still stresses the supply.
• Friction in efficiency calculations: Ignoring friction leads to an over‑estimate of efficiency; always include all energy losses.

Suggested Diagrams

Summary

Power quantifies how quickly work is done or energy is transferred. The fundamental definitions are

\[ P = \frac{dW}{dt} = \vec F\!\cdot\!\vec v = \frac{dE}{dt}, \]

which reduces to \(P = \dfrac{W}{t}\) for constant conditions. Mastery of the work‑energy theorem, the various mechanical and electrical forms of power, and the concept of efficiency equips students to tackle both theoretical questions and practical investigations required by the Cambridge AS & A‑Level Physics (9702) syllabus.