Power
How fast is work done?
Two students carry identical bags of books up the same staircase. One strolls up in a minute. The other runs up in fifteen seconds.
Both lift the same weight through the same height, so both do the same work. But the runner does it four times faster.
The rate of doing work is called power. Average power is the total work divided by the total time.
Instantaneous power is the rate at that moment: P equals dW by dt.
In a short time dt, a force F moves a body through dr, doing work F dot dr. Divide by dt and dr by dt becomes the velocity.
So P equals F dot v. A lift motor pulling with 10,000 newtons at 2 metres per second delivers 20,000 watts.
Like work, power is a scalar. Its unit, one joule per second, is called a watt.
Watts, horsepower and kilowatt-hours
The watt is named after James Watt, who improved the steam engine. A phone charger uses a few watts. A ceiling fan, about 75 watts.
Car and motorbike engines are often rated in horsepower. One horsepower is 746 watts.
Now look at your electricity bill. It does not charge you for power. It charges you for energy, in units.
One unit is one kilowatt-hour: the energy used by a 1000 watt appliance running for one hour.
That is 1000 joules every second for 3600 seconds: 3.6 million joules.
Careful: a kilowatt is a unit of power, but a kilowatt-hour is a unit of energy.
A 2 kilowatt geyser left on for half an hour uses 1 kilowatt-hour, which is one unit on the bill.
Try it: power of a lift motor
- Press Go up and watch the work–time line grow. Its slope is the power.
- Double the speed: the same work is done in half the time, so the power doubles.
- Add friction and see how much more force and power the motor needs.
Lift motor
A lift carrying a total load of 1800 kg (lift + passengers) moves up at a steady 2 m/s. Friction opposing the motion is 4000 N. What is the minimum power the motor must supply? (g = 9.8 m/s²)
- About 35 kW
- About 43 kW
- About 22 kW
- About 8 kW
Answer
B. At steady speed the motor’s force balances weight and friction: F = mg + f = 1800 × 9.8 + 4000 = 21 640 N. Then P = Fv = 21 640 × 2 ≈ 43 kW, about 58 hp. Forgetting friction gives about 35 kW. (NCERT uses g = 10 m/s² and gets 44 kW.)
Electricity units
A 1.5 kW immersion heater is used for 2 hours. How much energy does it use?
- 3 kWh, which is 1.08 × 10⁷ J
- 3 kW
- 0.75 kWh
- 3000 J
Answer
A. Energy = power × time = 1.5 kW × 2 h = 3 kWh (3 units on the bill). Since 1 kWh = 3.6 × 10⁶ J, this is 3 × 3.6 × 10⁶ = 1.08 × 10⁷ J. “3 kW” is a power, not an energy.
Summary
- Average power:
P_av = W / t - Instantaneous power:
P = dW/dt = F·v - Units:
1 W = 1 J s⁻¹, 1 hp = 746 W - Electrical energy:
1 kWh = 3.6 × 10⁶ J
- Power is the rate of doing work. The same work done faster needs more power.
- For a force moving a body at velocity v, P = F·v. Power is a scalar.
- The kilowatt-hour is a unit of energy, not power. It is the “unit” on your electricity bill.