Power and efficiency IB DP Physics Study Notes - 2025 Syllabus
Power and efficiency IB DP Physics Study Notes
Power and efficiency IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
- that power developed $P$ is the rate of work done, or the rate of energy transfer, as given by
$P = \frac{\Delta W}{\Delta t} = Fv$ - efficiency $\eta$ in terms of energy transfer or power as given by
$\eta = \frac{E_{\text{output}}}{E_{\text{input}}} = \frac{P_{\text{output}}}{P_{\text{input}}}$
Standard level and higher level:8 hours
Additional higher level: There is no additional higher level content
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Power
In physics, power is defined as the rate at which work is done or energy is transferred.
The SI unit of power is the watt (W), where:
- \( 1 \, \text{Watt} = 1 \, \text{Joule/second} \)
There are two commonly used formulas for power:
- Average Power: \( P = \frac{W}{t} \), where \( W \) is work done and \( t \) is time taken
- Instantaneous Power: \( P = Fv \cos \theta \), where \( F \) is force, \( v \) is velocity, and \( \theta \) is the angle between force and velocity vectors
Power can also describe:
- How fast a machine does work
- How quickly energy is converted (e.g. in electric circuits)
Example:
A motor lifts a \( 100 \, \text{kg} \) mass vertically upward at a constant speed to a height of \( 5.0 \, \text{m} \) in \( 4.0 \, \text{s} \). Find the power developed by the motor.
▶️ Answer/Explanation
Step 1: Calculate work done against gravity
Work done = \( W = mgh = 100 \cdot 9.8 \cdot 5 = 4900 \, \text{J} \)
Step 2: Use power formula
\( P = \frac{W}{t} = \frac{4900}{4.0} = \boxed{1225 \, \text{W}} \)
Interpretation:
The motor converts electrical energy into gravitational potential energy at a rate of \( 1225 \, \text{W} \).
Efficiency
Efficiency \( (\eta) \) is the ratio of useful energy output to total energy input. It tells us how effectively energy is transformed or transferred in a system.
The formula for efficiency in terms of energy is:
\( \eta = \frac{\text{Useful Output Energy}}{\text{Input Energy}} \)
The formula in terms of power is:
\( \eta = \frac{\text{Useful Power Output}}{\text{Power Input}} \)
To express efficiency as a percentage:
\( \eta = \left( \frac{\text{Useful Output Energy}}{\text{Input Energy}} \right) \times 100\% \quad \text{or} \quad \eta = \left( \frac{\text{Useful Power Output}}{\text{Power Input}} \right) \times 100\% \)
Efficiency is always less than 100% in real systems due to energy losses (e.g., heat, sound, vibration).
Example:
An electric motor consumes \( 1200 \, \text{W} \) and provides a useful mechanical output of \( 900 \, \text{W} \). Calculate its efficiency using \( \eta \).
▶️ Answer/Explanation
Step 1: Use power-based efficiency formula
\( \eta = \frac{900}{1200} = 0.75 \quad \Rightarrow \quad \boxed{\eta = 75\%} \)
Interpretation:
75% of the input energy is converted to useful output; the rest is lost to surroundings.