Contents
Key Stage 3
Meaning
Power is the rate of Energy Transfer.
About Power
- The unit of power is the Watt.
- Power is how quickly energy is transferred from one store into another.
- Energy Transfered is the same as Work Done, so power is also the rate of work done.
Equation
\(Power = \frac{Energy Transferred}{Time}\)
\(P = \frac{E}{t}\)
\(Power = \frac{Work Done}{Time}\)
\(P = \frac{W}{t}\)
Where:
P = Power
E = Energy Transferred = W = Work Done
t = Time
Key Stage 4
Meaning
Power is the rate of Energy Transfer or Work Done.
About Power
- The SI Unit of power is the watt.
- Power is a scalar quantity as it has magnitude but does not have a direction.
Equations
Power, Work Done and Time
Power = \(\frac{Work Done}{Time}\)
\[P = \frac{W}{t}\]
Where:
P = Power
E = Energy Transferred = W = Work Done
t = Time
Power, Current and Potential Difference
Power = (Current) x (Potential Difference)
\[P = IV\]
Where:
P = Power
I = Current
Calculating Power from Work Done
| A bow at full stretch transfers 50J of energy from its elastic potential energy store into the kinetic energy store of an arrow in 0.02 seconds. Calculate the power of this transfer correct to two significant figures. | A piano falls down a flight of stairs. In this fall 5.4kJ of energy is transferred from the gravitational potential energy store into the kinetic energy store of the piano. It take 0.80 seconds to reach the bottom of the stairs. Calculate the average power of this transfer correct to two significant figures. | A 'weight lifter' lifts a 2000N weight a distance of 1.2 metre from its original position. They do this in 0.70 seconds. Calculate the power output of the 'weight lifter' correct to two significant figures. |
| 1. State the known quantities
W = 50J t = 0.02s |
1. State the known quantities
W = 5.4kJ = 5400J t = 0.80s |
1. State the known quantities
F = 2000N d = 1.2m t = 0.70s |
| 2. Substitute the numbers into the equation and solve.
\(P = \frac{W}{t}\) \(P = \frac{50}{0.02}\) \(P = 2500W\) |
2. Substitute the numbers into the equation and solve.
\(P = \frac{W}{t}\) \(P = \frac{5400}{0.80}\) \(P = 6750W\) \(P \approx 6800W\) |
2. Substitute the numbers into the equation and solve.
\(W = \vec F \vec d\) \(W = \vec F \times \vec d\) \(W = 2000 \times 1.2\) \(W = 2400J\) \(P = \frac{W}{t}\) \(P = \frac{2400}{0.7}\) \(P = 3428.57W\) \(P \approx 3400W\) |
Calculating Electrical Power
| An electrical fan is plugged into a socket supplying 230V. An ammeter reading shows that the current going into the fan is 0.20amps. Calculate the power of the fan correct to two significant figures. | A light bulb is supplied a current of 625mA. The potential difference across the terminals on the light bulb is 96 volts. Calculate the power of the bulb correct to two significant figures. | A kettle is plugged into the mains supply which operates at 230V. It receives a current of 9.2Amps Calculate the power of the kettle correct to two significant figures. |
| 1. State the known quantities
I = 0.20A V = 230V |
1. State the known quantities
I = 625mA = 0.625A V = 96V |
1. State the known quantities
I = 9.2A V = 230V |
| 2. Substitute the numbers into the equation and solve.
\(P = IV\) \(P = I \times V\) \(P = 0.20 \times 230\) \(P = 46W\) |
2. Substitute the numbers into the equation and solve.
\(P = IV\) \(P = I \times V\) \(P = 0.625 \times 96\) \(P = 60W\) |
2. Substitute the numbers into the equation and solve.
\(P = IV\) \(P = I \times V\) \(P = 9.2 \times 230\) \(P = 2116W\) \(P \approx 2100W\) |