Power Knowledge-Based Study

Study notes

1. What is Power?

   In physics, power is defined as the rate at which work is done or the rate at which energy is transferred or converted from one form to another.

   While work measures the total amount of energy transferred, power indicates how rapidly that transfer takes place. If two machines perform the same total amount of work, the machine that completes it in a shorter amount of time delivers greater power.

   The mathematical definition of average power is:

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

   • P: Power
   • W: Work done (or \Delta E, energy transferred)
   • t: Time elapsed

   Units and Conversions

   The standard SI unit of power is the watt (symbol: W), named in honor of the Scottish engineer James Watt.

   One watt corresponds to one joule of work completed per second:

   $$1\text{ W} = 1\text{ J/s} = 1\text{ kg}\cdot\text{m}^2/\text{s}^3$$

   Common multiple units include:

   • Kilowatt (kW): $1\text{ kW} = 10^3\text{ W}$
   • Megawatt (MW): $1\text{ MW} = 10^6\text{ W}$
   • Horsepower (hp): An imperial unit often used in automotive and mechanical contexts, where $1\text{ hp} \approx 746\text{ W}$.

   Relating Power to Motion

   When a constant net force $F$ acts on an object moving along the direction of the force with a constant displacement $d$ over time interval $t$, we can express power as:

   $$P = \frac{W}{t} = \frac{F \cdot d}{t}$$

   Since average speed or velocity is given by $v = \frac{d}{t}$, this yields:

   $$P = F \cdot v$$

   If the force is applied at an angle $\theta$ to the direction of motion, the instantaneous power delivered is:

   $$P = F v \cos(\theta)$$

   This formulation demonstrates that maintaining a higher speed against a resisting force (such as friction or aerodynamic drag) requires proportionately greater power output.

   Calculating Power in Practice

   Consider an electric motor lifting an object of mass $m$ through a vertical height $h$ at a constant speed over a time duration $t$.

   • The work done against gravity is: $$W = mgh$$
   • The power output required by the motor is: $$P = \frac{mgh}{t}$$

   Example: A crane lifts a load of 500 kg vertically by 12 meters in 6 seconds at constant speed. Taking $g = 9.8\text{ m/s}^2$:

   $$W = 500 \times 9.8 \times 12 = 58800\text{ J}$$

   $$P = \frac{58800\text{ J}}{6\text{ s}} = 9800\text{ W} = 9.8\text{ kW}$$

   If the same task were performed in 3 seconds instead of 6 seconds, twice as much power (19.6 kW) would be required, even though the total work performed remains identical.