Theory of Intelligence
11TH•Physics•Work Energy And Power•Power

4.4 Power

Rate of doing work, P = F · v, the watt, and kilowatt-hour.

4.4 Power

In the definition of work, it is not clear, whether the same amount of work is done in one second or in one hour. The rate, at which work is done, is often of interest in practical applications.

Power is the measure of the rate at which work is being done.

If work ΔW\Delta W is done in a time interval Δt\Delta t, then the average power PavP_{\mathrm{av}} during the interval Δt\Delta t is defined as:

Pav=ΔWΔt…(4.6)P_{\mathrm{av}} = \frac{\Delta W}{\Delta t} \quad \ldots (4.6)

If work is expressed as a function of time, the instantaneous power PP at any instant is defined as:

P=lim⁡Δt→0ΔWΔt…(4.7)P = \lim_{\Delta t \to 0} \frac{\Delta W}{\Delta t} \quad \ldots (4.7)

where ΔW\Delta W is the work done in short interval of time Δt\Delta t.

Since ΔW=F⋅Δd\Delta W = \mathbf{F}\cdot\Delta\mathbf{d}

Hence P=(F⋅Δd)/Δt=F⋅(Δd/Δt)P = (\mathbf{F}\cdot\Delta\mathbf{d})/\Delta t = \mathbf{F}\cdot(\Delta\mathbf{d}/\Delta t)

Since Δd/Δt=v\Delta\mathbf{d}/\Delta t = \mathbf{v}

Hence

P=F⋅v…(4.8)P = \mathbf{F}\cdot\mathbf{v} \quad \ldots (4.8)

Thus, power may also be defined as the scalar product of F\mathbf{F} and v\mathbf{v}.

The SI unit of power is watt, defined as one joule of work done in one second.

Sometimes, for example, in the electrical measurements, the unit of work is expressed as watt second. However, a commercial unit of electrical energy is kilowatt-hour.

One kilowatt-hour is the work done in one hour by an agency whose power is one kilowatt.

Therefore

1 kW⋅h=1000 W×3600 s1\,\mathrm{kW}\cdot\mathrm{h} = 1000\,\mathrm{W} \times 3600\,\mathrm{s}

or

1 kW⋅h=3.6×106 J=3.6 MJ1\,\mathrm{kW}\cdot\mathrm{h} = 3.6 \times 10^{6}\,\mathrm{J} = 3.6\,\mathrm{MJ}

For your information — Approximate Power

DeviceApproximate Power (W)
Jumbo Jet Aircraft1.3×1041.3 \times 10^{4}
Car at 90 km/h90\,\mathrm{km/h}1.1×1031.1 \times 10^{3}
Electric heater2×1032 \times 10^{3}
Coloured TV120120

Example 4.2

A 70 kg70\,\mathrm{kg} man runs up a long flight of stairs in 4.0 s4.0\,\mathrm{s}. The vertical height of the stairs is 4.5 m4.5\,\mathrm{m}. Calculate his power output in watts.

Solution

Work done =mgh= mgh

Power P=mgh/tP = mgh/t

P=(70 kg×9.8 m/s2×4.5 m)4 sP = \frac{(70\,\mathrm{kg} \times 9.8\,\mathrm{m/s^{2}} \times 4.5\,\mathrm{m})}{4\,\mathrm{s}} P=7.7×103 kg m/s3=7.7×103 W\mathbf{P = 7.7 \times 10^{3}\,\mathrm{kg\,m/s^{3}} = 7.7 \times 10^{3}\,\mathrm{W}}

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