Edexcel A Level (IAL) Physics-1.21 -1.22 Power & Efficiency- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -1.21 -1.22 Power & Efficiency- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -1.21 -1.22 Power & Efficiency- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 1.21 be able to use the equations relating power, time and energy transferred or work done $P = \frac{E}{t}$ and $P = \frac{W}{t}$
- 1.22 be able to use the equations $\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}$ and $\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}}$
Power, Energy, Work and Time \( P = \dfrac{E}{t} \) and \( P = \dfrac{W}{t} \)
Definition of Power
Power is the rate at which work is done or the rate at which energy is transferred.
\( P = \dfrac{E}{t} \quad \text{and} \quad P = \dfrac{W}{t} \)
- \( P \) = power (W)
- \( E \) = energy transferred (J)
- \( W \) = work done (J)
- \( t \) = time taken (s)
Key Points
- 1 watt (W) = 1 joule per second.
- If more energy is transferred in a shorter time → higher power.
- If a machine has high power, it can do work quickly.
- Power can also be calculated using force and velocity:
\( P = Fv \)
Understanding the Equations
- Use \( P = \dfrac{E}{t} \) when energy transfer is known (e.g., electrical power).
- Use \( P = \dfrac{W}{t} \) when mechanical work is involved (lifting, pulling, pushing).
- If work is done at an angle, use the component of force along the displacement.
- Power can be negative if the force removes energy (e.g., brakes).
Example (Easy)
A motor transfers \( 300\, \mathrm{J} \) of energy in \( 10\, \mathrm{s} \). Find its power output.
▶️ Answer / Explanation
\( P = \dfrac{E}{t} = \dfrac{300}{10} = 30\, \mathrm{W} \)
Example (Medium)
A person pushes a box with a force of \( 60\, \mathrm{N} \) at a constant speed of \( 2.5\, \mathrm{m\,s^{-1}} \). Find the power delivered to the box.
▶️ Answer / Explanation
Use \( P = Fv \):
\( P = 60 \times 2.5 = 150\, \mathrm{W} \)
Example (Hard)
An elevator lifts a \( 500\, \mathrm{kg} \) load vertically through \( 12\, \mathrm{m} \) in \( 8\, \mathrm{s} \). Calculate the power needed (ignore friction).
▶️ Answer / Explanation
Work done = gain in GPE:
\( W = mg\Delta h = 500 \times 9.8 \times 12 = 58800\, \mathrm{J} \)
Power:
\( P = \dfrac{W}{t} = \dfrac{58800}{8} = 7350\, \mathrm{W} \)
The elevator requires a power of \( 7.35\, \mathrm{kW} \).
Efficiency Using Energy and Power
Definition of Efficiency
Efficiency measures how much of the input energy or power is usefully transferred. It is always a number between 0 and 1 (or 0% and 100%).
Efficiency in Terms of Energy:
\( \text{Efficiency} = \dfrac{\text{Useful Energy Output}}{\text{Total Energy Input}} \)
Efficiency in Terms of Power:
\( \text{Efficiency} = \dfrac{\text{Useful Power Output}}{\text{Total Power Input}} \)
- Efficiency has no units.
- Multiply by 100 to express as a percentage.
- Useful output ≤ total input (always).
- Energy lost is usually wasted as heat, sound, or vibration.
Key Ideas
- Perfect efficiency = 1 (or 100%) — impossible in real machines.
- Higher efficiency → less energy wasted.
- Low efficiency → more energy lost to surroundings.
- Use energy formula for processes; use power formula for machines running continuously.
Example (Easy)
A machine uses \( 200\, \mathrm{J} \) of energy and delivers \( 150\, \mathrm{J} \) of useful energy. Find its efficiency.
▶️ Answer / Explanation
\( \text{Efficiency} = \dfrac{150}{200} = 0.75 = 75\% \)
Example (Medium)
A motor takes in \( 500\, \mathrm{W} \) of power and produces \( 350\, \mathrm{W} \) of useful mechanical power. Calculate its efficiency.
▶️ Answer / Explanation
\( \text{Efficiency} = \dfrac{350}{500} = 0.70 = 70\% \)
Example (Hard)
An engine receives \( 2.4 \times 10^{6}\, \mathrm{J} \) of energy each minute. In the same time it produces \( 1.5 \times 10^{6}\, \mathrm{J} \) of useful energy and wastes the rest as heat. Find:
- (a) Its energy efficiency
- (b) Its useful power output
▶️ Answer / Explanation
(a) Energy Efficiency
\( \text{Efficiency} = \dfrac{1.5\times 10^6}{2.4\times 10^6} = 0.625 = 62.5\% \)
(b) Useful Power Output
Time = 1 minute = 60 s.
\( P = \dfrac{E}{t} = \dfrac{1.5\times10^6}{60} = 2.5\times 10^4\, \mathrm{W} \)
Useful power = \( 25\, \mathrm{kW} \).