define power as work done per unit time

Energy Conservation

Objective

Define power as the rate at which work is done (or energy is transferred) per unit time.

Key Concepts

• Work ($W$): The product of a force component parallel to the displacement and the displacement itself. $$W = \int \vec{F}\cdot d\vec{s}$$
• Energy ($E$): The capacity to do work. In mechanics, the primary forms are kinetic and potential energy.
• Power ($P$): The rate of doing work or transferring energy. $$P = \frac{dW}{dt} = \frac{dE}{dt}$$

Definition of Power

Power is defined as the amount of work done per unit time. In symbols:

$$P = \frac{W}{t}$$

When the work is done at a constant rate, the above expression is sufficient. For varying forces or speeds, the instantaneous power is given by the derivative:

$$P = \frac{dW}{dt} = \vec{F}\cdot\vec{v}$$

where $\vec{v}$ is the instantaneous velocity of the point of application of the force.

Units

Quantity	SI Unit	Symbol
Work / Energy	joule	J
Power	watt	W
Time	second	s

Since $1\ \text{W} = 1\ \text{J}\,\text{s}^{-1}$, power can also be expressed in other units, e.g., $1\ \text{kW}=1000\ \text{W}$.

Relationship to Energy Conservation

The principle of energy conservation states that the total energy of an isolated system remains constant. Power provides a way to quantify how quickly energy is transferred or transformed within the system.

1. Identify all forms of energy in the system (kinetic, gravitational potential, elastic, etc.).
2. Write the energy balance: $$\Delta E_{\text{total}} = 0$$ for an isolated system.
3. Differentiate with respect to time to obtain a power balance: $$\frac{dE_{\text{total}}}{dt}=0$$ which implies that the sum of all power inputs equals the sum of all power outputs.

Example Problem

Problem: A 1500 kg car accelerates from rest to $20\ \text{m s}^{-1}$ in $10\ \text{s}$. Assuming the engine provides a constant force, calculate the average power delivered by the engine.

Solution:

1. Calculate the work done (change in kinetic energy): $$\Delta K = \frac{1}{2}mv^{2} = \frac{1}{2}(1500\ \text{kg})(20\ \text{m s}^{-1})^{2}=3.0\times10^{5}\ \text{J}$$
2. Average power: $$P_{\text{avg}} = \frac{\Delta K}{\Delta t}= \frac{3.0\times10^{5}\ \text{J}}{10\ \text{s}} = 3.0\times10^{4}\ \text{W}=30\ \text{kW}$$

Common Misconceptions

• Confusing energy (J) with power (W). Energy is a quantity; power is its rate of change.
• Assuming power is always constant; in many real situations it varies with time.
• Neglecting the vector nature of force and velocity when using $P = \vec{F}\cdot\vec{v}$.

Suggested Diagram

Summary

Power quantifies how quickly work is done or energy is transferred. It is defined as $P = \frac{dW}{dt}$ and, for constant forces, simplifies to $P = \frac{W}{t}$. Understanding power is essential for applying the principle of energy conservation to dynamic systems.