Work=Force∗Distance{\displaystyle {\text{Work}}={\text{Force}}*{\text{Distance}}} To understand this concept, think about a 100 pound cement block. Moving that block 2 feet requires twice as much work as moving the same block 1 foot. This is also the same amount of work required to move a 200 pound block for a distance of 6 inches (1/2 a foot).
Power=WorkTime{\displaystyle {\text{Power}}={\frac {\text{Work}}{\text{Time}}}} Think again about the 100 pound cement block. Moving that block 2 feet in 10 seconds requires twice as much power as moving the same block for the same distance in 20 seconds. Moving the same object faster requires more power.
The English unit for work is the “foot-pound. ” This comes from multiplying the distance measured in feet by the force measured in pounds. The unit for power comes from dividing by time, in seconds. The power unit is the “foot-pound per second. ” A more familiar term is the “horsepower,” which is defined as 550 foot-pounds per second. [3] X Research source 1 Horsepower=550foot-poundssec. {\displaystyle 1{\text{ Horsepower}}=550{\frac {\text{foot-pounds}}{\text{sec. }}}} The metric unit for work is a joule. A joule is equal to one Newton-meter, which is the amount of work needed to move a 1 Newton force a distance of 1 meter. When you divide work by time, you get power, measured in units of joules per second. This is also called a Watt. [4] X Research source 1 Watt=1 Joule Sec. =1 Newton-Meter Sec. {\displaystyle 1{\text{ Watt}}=1{\frac {\text{ Joule}}{\text{ Sec. }}}=1{\frac {\text{ Newton-Meter}}{\text{ Sec. }}}}
For example, suppose you want to calculate in the following problem: A student doing chin-ups lifts his 150 lb. body a distance of 1. 5 feet in 2 seconds. How much power is the student exerting? Force=150 lbs. {\displaystyle {\text{Force}}=150{\text{ lbs. }}} Distance=1. 5 feet{\displaystyle {\text{Distance}}=1. 5{\text{ feet}}} Time=2 sec. {\displaystyle {\text{Time}}=2{\text{ sec. }}}
Power=Force∗DistanceTime{\displaystyle {\text{Power}}={\frac {{\text{Force}}*{\text{Distance}}}{\text{Time}}}} Power=150lb. ∗1. 5feet2sec. {\displaystyle {\text{Power}}={\frac {150{\text{lb. }}*1. 5{\text{feet}}}{2{\text{sec. }}}}} Power=225ft-lb2sec. {\displaystyle {\text{Power}}={\frac {225{\text{ft-lb}}}{2{\text{sec. }}}}} Power=112. 5ft-lbsec. {\displaystyle {\text{Power}}=112. 5{\frac {\text{ft-lb}}{\text{sec. }}}} Since 1 horsepower equals 550 ft-lb/sec. , this is about equal to one-fifth of a horsepower.
Power=Force∗DistanceTime{\displaystyle {\text{Power}}={\text{Force}}{\frac {\text{Distance}}{\text{Time}}}} By grouping the terms differently, in this way, you should recognize that DistanceTime{\displaystyle {\frac {\text{Distance}}{\text{Time}}}} is another way of representing velocity. Therefore, you can conclude that: Power=Force∗Velocity{\displaystyle {\text{Power}}={\text{Force}}{\text{Velocity}}} Consider a 2000 pound car. Moving that car at a speed of 50 miles per hour takes twice as much power as moving the same car at a speed of 25 miles per hour.
1 mile = 5280 feet 1 hour = 60 minutes = 3600 seconds 1 kg = 9. 8 N (force) For the moving car problem, the data is provided in miles, hours and pounds. You need to convert 50 miles into 50*5280=264,000 feet, and convert hours into 3600 seconds. Then you can restate the problem as calculating the power of moving a 2000 pound car at a velocity of 264,000/3,600 = 73. 33 feet per second.
Power=Force∗Velocity{\displaystyle {\text{Power}}={\text{Force}}*{\text{Velocity}}} Power=2000Pounds∗73. 33feetsec{\displaystyle {\text{Power}}=2000{\text{Pounds}}*73. 33{\frac {\text{feet}}{\text{sec}}}} Power=146,660foot-poundssec{\displaystyle {\text{Power}}=146,660{\frac {\text{foot-pounds}}{\text{sec}}}} You can report this as the car’s power, or convert to horsepower by dividing by 550. Power=146,660/550=266. 65 horsepower{\displaystyle {\text{Power}}=146,660/550=266. 65{\text{ horsepower}}} The engine exerted 266. 65 horsepower in moving the 2000 pound car at a velocity of 50 mph.
If you are reading a circuit diagram, the voltage will be written somewhere on the diagram. If you are building an actual circuit, the voltage is the power coming from the battery source. For example, a single 9 volt battery provides 9 volts to the circuit.
Current can be measured directly using a reader called an ammeter. [12] X Research source Alternatively, the problem you are working on may simply provide the number of Amps in the circuit. For example, a circuit that has a current of 4 Amps is moving the charge twice as quickly as a circuit with a current only 2 Amps.
The unit of electrical power is the Watt. It is calculated as follows: Power=Current∗Voltage{\displaystyle {\text{Power}}={\text{Current}}*{\text{Voltage}}} P=IE{\displaystyle P=IE}
Power=Current∗Voltage{\displaystyle {\text{Power}}={\text{Current}}{\text{Voltage}}} Power=6∗18=108Watts{\displaystyle {\text{Power}}=618=108Watts}
