Power and Efficiency in Physics

In everyday language, "power" often means strength, but in physics it has a precise meaning: power is the rate at which work is done, or the rate at which energy is transferred. Two machines can do exactly the same amount of work, but the one that finishes faster is more powerful. If you have already studied work and energy, power simply adds a time element to that idea.

The basic formula for power is:

\( P = \dfrac{W}{t} \)

where \(P\) is power in watts (W), \(W\) is the work done in joules (J), and \(t\) is the time taken in seconds (s). One watt is equal to one joule of work done per second (\(1\ \)W\( = 1\ \)J/s\(\)).

Since work done by a constant force is \(W = Fd\), where \(d\) is the distance moved in the direction of the force, we can substitute this into the power formula:

\( P = \dfrac{Fd}{t} \)

Because \(\dfrac{d}{t}\) is velocity \(v\), this simplifies to a second useful power equation:

\( P = Fv \)

This version is especially handy when you know the force acting on an object and the speed at which it moves, such as a car engine pushing a vehicle forward at a steady speed. The graph below shows how power changes with velocity when the force stays constant at 50 N.

A crane lifts a 500 kg crate a height of 10 m in 5 seconds. Find the power output of the crane.

Step 1: Find the work done against gravity.

\( W = mgh = (500)(9.8)(10) = 49{,}000\ \)J\( \)

Step 2: Divide by the time taken.

\( P = \dfrac{W}{t} = \dfrac{49{,}000}{5} = 9{,}800\ \)W\( = 9.8\ \)kW\( \)

The crane delivers 9.8 kilowatts of useful power while lifting the crate. This calculation relies directly on the work and energy relationships covered when studying conservation of energy, since the work done against gravity becomes stored gravitational potential energy.

No real machine converts all the energy it receives into useful work. Some energy is always lost, usually as heat due to friction, air resistance, or internal resistance in motors. Mechanical efficiency measures how much of the input energy actually becomes useful output. It is expressed as a percentage:

\( \)Efficiency\( = \dfrac{\)Useful energy output\(}{\)Total energy input\(} \times 100\% \)

Because power is just energy transferred per unit time, the same ratio also works directly with power values, as long as the input and output are measured over the same time interval:

\( \)Efficiency\( = \dfrac{\)Useful power output\(}{\)Total power input\(} \times 100\% \)

An efficiency of 100% is impossible for any real mechanical system, since some energy is always lost to heat, sound, or friction.

An electric motor draws 12,000 W of electrical power to lift a crate. The useful mechanical power output, from the previous example, is 9,800 W. Find the efficiency of the motor.

\( \)Efficiency\( = \dfrac{9{,}800}{12{,}000} \times 100\% \approx 81.7\% \)

This means about 81.7% of the electrical energy supplied to the motor is converted into useful work lifting the crate, while the remaining 18.3% is lost, mostly as heat inside the motor's moving parts.

Power tells you how quickly energy is being transferred or work is being done, while efficiency tells you how much of that transferred energy is actually useful. A machine can be very powerful but still inefficient if most of its output power is wasted as heat. When solving problems, first identify whether you need a rate (power), a ratio (efficiency), or both, then choose the matching formula: \(P = \dfrac{W}{t}\), \(P = Fv\), or \(\)Efficiency\( = \dfrac{\)output\(}{\)input\(} \times 100\%\).

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