Average Power — Revision Notes

⚡ 30-Second Revision

Definition: — $P_{avg} = \frac{\Delta W}{\Delta t} = \frac{\Delta E}{\Delta t}$
Units: — Watt (W) = Joule/second (J/s). $1,\text{hp} \approx 746,\text{W}$.
Constant Force: — $P_{avg} = F \cdot v_{avg}$ (if force is constant and in direction of motion).
Work Done: — $W = Fd\cos\theta$, $W = mgh$, $W = \frac{1}{2}mv^2$, $W = \int F(x) dx$, $W = \int P(t) dt$.
Efficiency: — $\eta = \frac{P_{out}}{P_{in}} = \frac{\text{Useful Power Output}}{\text{Total Power Input}}$.
Graphical: — Area under $P-t$ graph gives total work/energy.

2-Minute Revision

Average power is the overall rate at which work is done or energy is transferred over a specific time interval. It's calculated as the total work done ($\Delta W$) divided by the total time taken ($\Delta t$), i.

e., $P_{avg} = \frac{\Delta W}{\Delta t}$. The SI unit is the Watt (W), equivalent to $1,\text{J/s}$. This concept is crucial when the rate of work is not constant, providing a smoothed-out value. For a constant force, average power can also be expressed as the product of the force and the average velocity ($P_{avg} = F \cdot v_{avg}$).

Remember that total work can be found from changes in kinetic or potential energy, or by integrating a variable force over displacement ($W = \int F(x) dx$). If instantaneous power $P(t)$ is given, total work is $\int P(t) dt$.

Efficiency problems are common, where $\eta = \frac{P_{out}}{P_{in}}$, requiring you to distinguish between useful power output and total power input. Always pay attention to units and ensure consistency.

5-Minute Revision

Average power is a measure of how quickly work is performed or energy is transferred over a finite period. It's fundamentally defined as $P_{avg} = \frac{\text{Total Work Done}}{\text{Total Time Taken}} = \frac{\Delta W}{\Delta t}$. This is particularly useful when the force, velocity, or rate of energy transfer varies over time, giving us an overall picture rather than a momentary one. The SI unit is the Watt (W), where $1,\text{W} = 1,\text{J/s}$.

Key Formulas and Concepts:

1
Work-Energy Principle: — Total work done on an object equals its change in kinetic energy, $W = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$. This is often the starting point for calculating work.
2
Work against Gravity: — For lifting an object of mass $m$ to height $h$, $W = mgh$.
3
Work against Friction: — $W_f = f_k d = \mu_k N d$.
4
Variable Force: — If force $F$ is a function of position $x$, $W = \int_{x_i}^{x_f} F(x) dx$.
5
Average Power from Force and Average Velocity: — If a constant force $F$ acts on an object moving with average velocity $v_{avg}$ in the direction of the force, then $P_{avg} = F \cdot v_{avg}$. Remember $v_{avg} = \frac{\Delta x}{\Delta t}$.
6
Efficiency ($\eta$): — For any machine, $\eta = \frac{\text{Power Output}}{\text{Power Input}} = \frac{P_{out}}{P_{in}}$. This is crucial for problems involving pumps, motors, etc., where some energy is always lost.
7
Power-Time Graphs: — The area under a power-time ($P-t$) graph represents the total work done or energy transferred. If $P(t)$ is given, $\Delta W = \int P(t) dt$.

Example: A $100,\text{kg}$ block is pulled $5,\text{m}$ up a $30^circ$ incline in $2,\text{s}$ at constant velocity. Coefficient of friction is $0.2$. Calculate average power.

Solution:

$F_g = mg\sin\theta = 100 \times 10 \times 0.5 = 500,\text{N}$.
$N = mg\cos\theta = 100 \times 10 \times \frac{\sqrt{3}}{2} = 500\sqrt{3},\text{N}$.
$f_k = \mu_k N = 0.2 \times 500\sqrt{3} = 100\sqrt{3},\text{N} \approx 173.2,\text{N}$.
$F_{pull} = F_g + f_k = 500 + 173.2 = 673.2,\text{N}$.
Work done $W = F_{pull} \times d = 673.2,\text{N} \times 5,\text{m} = 3366,\text{J}$.
Average Power $P_{avg} = \frac{W}{t} = \frac{3366,\text{J}}{2,\text{s}} = 1683,\text{W} = 1.683,\text{kW}$.

Remember to always convert units to SI, distinguish between average and instantaneous power, and carefully identify all forces and energy changes involved.

Prelims Revision Notes

Average power is the rate of doing work or transferring energy over a finite time interval. It is defined as $P_{avg} = \frac{\Delta W}{\Delta t}$ or $P_{avg} = \frac{\Delta E}{\Delta t}$. The SI unit is the Watt (W), where $1,\text{W} = 1,\text{J/s}$. Other units include horsepower ($1,\text{hp} \approx 746,\text{W}$). This concept is crucial for understanding the overall performance of systems where the instantaneous power might vary.

Key Formulas and Relationships:

Work Done:

* By constant force: $W = Fd\cos\theta$. * Against gravity: $W = mgh$. * Change in kinetic energy: $W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$. * By variable force: $W = \int F(x) dx$. * From instantaneous power: $W = \int P(t) dt$.

Average Velocity: — $v_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}}$.
Average Power (constant force): — $P_{avg} = F \cdot v_{avg}$ (if force is constant and parallel to average velocity).
Efficiency: — $\eta = \frac{P_{out}}{P_{in}} = \frac{\text{Useful Power Output}}{\text{Total Power Input}}$. Remember $\eta$ is unitless and typically expressed as a percentage or a decimal between 0 and 1.

Common Traps:

Confusing average power with instantaneous power. Instantaneous power is $P = \vec{F} \cdot \vec{v}$.
Confusing power with energy. Power is the *rate* of energy transfer.
Incorrectly applying efficiency (e.g., multiplying by efficiency instead of dividing when finding input power from output power).
Errors in unit conversions.
Not accounting for all forces (like friction) when calculating total work done.

Strategy: For numerical problems, first identify the total work done or energy change, then the total time. For variable quantities, integration is often required. For efficiency problems, clearly distinguish between input and output power. Always ensure consistent units throughout the calculation.

Vyyuha Quick Recall

Work-Energy-Time: What Energy Transfers? Per Time! (P = W/T or E/T)