Power

Show/Hide Sub-topics (Work, Energy & Power | O Level Physics)
Show/Hide Sub-topics (Work, Energy and Power | A Level Physics)

Power

Power is defined as the rate of work done or energy converted with respect to time.

$P = \frac{W}{t}$ OR $P = \frac{E}{t}$

, where W = work, t = time, E = Energy

SI Unit for power is watt (W), scalar quantity

• One watt (W) is defined as the rate of work done or energy conversion of one joule per second.
• $\text{One watt} = \frac{\text{one joule}}{\text{one second}}$ $\rightarrow$ $1 \, \text{W} = 1 \, \text{J s}^{-1}$

Power tells us how fast work is being done or how fast energy is being converted from one form to another.

Power In Terms Of Force & Velocity

An useful equation of power is as follows:

$$P = Fv$$

, where F = force, v = velocity (Simple derivation below)

Simple Derivation Of P = $Fv$

In some questions, the formulation of power in terms of force and velocity will be useful in the problem-solving. The formula $P = Fv$ can be simply derived as seen below:

\begin{aligned} P &= \frac{W}{t} \\ &= \frac{F \times d}{t} \\ &= Fv \text{ where } v = \frac{d}{t} \end{aligned}

Efficiency

From the Principle of Conservation of Energy, we know that the total energy output of a machine must be equal to its energy input. However, it is found that the energy output of a machine is always less than the energy input. This phenomena can be attributed to the work done against frictional forces, which is considered as wasted energy output.

Hence, we have:

Energy input = useful energy output + wasted energy output

Efficiency of a system is given by

$$\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100 \%$$

Worked Examples

Example 1: Fat Man & Thin Man

A fat man and a thin man (with half of the mass of the fat man) ran to the top of a hill in the same time. Whose power output is higher? Why?

Recall that power is given by $P = \frac{W}{t}$.

Notice that the time, t, is the same for both fat man and thin man. Hence, we will have to look at the work done by both men.

The work done by the fat man will be larger than the work done by the thin man.

Hence, the power output by the fat man is higher.

Example 2: Power of engine

An engine does 60 000 J of work in ten minutes. What is the power of the engine?

\begin{aligned} P &= \frac{W}{t} \\ &= \frac{60 000 \, \text{J}}{10 \times 60 \, \text{s}} \\ &= 100 \, \text{W} \end{aligned}

Example 3: Efficiency of Electric Motor

An electric motor is rated at 1.0 kW. If 60% of the input energy is lost as heat and sound, find the amount of useful energy produced in half hour. Hence or otherwise, find the efficiency of the electric motor.

Let’s begin by calculating the total energy output of the electric motor:

\begin{aligned} P &= \frac{W}{t} \\ W &= P \times t \\ &= 1000 \, \text{W} \times 30 \times 60 \, \text{s} \\ &= 1.8 \times 10^{6} \, \text{J} \end{aligned}

Since 60% of the energy from the electric motor is lost as heat and sound, only 40% is useful energy output. Hence,

\begin{aligned} \text{Useful Energy Output} &= \frac{40}{100} \times 1.8 \times 10^{6} \\ &= 7.2 \times 10^{5} \, \text{J} \end{aligned}

As given in the question, the efficiency of the electric motor is 40%.

Example 4: Electric Motor

An electric motor is used to lift a 10 N load through 5 m. The total amount of electrical energy input is 65 J. Calculate the amount of energy wasted by the motor. Hence or otherwise, calculate the efficiency of the motor.

Since efficiency is given by: $\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100 \%$, we have to find the useful energy output. In this case, the useful energy output is the lifting of the 10 N load through 5 m.

Hence, useful energy output:

\begin{aligned} W_{\text{useful}} &= F \times d \\ &= 10 \times 5 \\ &= 50 \text{ J} \end{aligned}

Since the total amount of electrical energy input is given to be 65 J, the energy wasted will be:

\begin{aligned} E_{\text{wasted}} &= 65-50 \\ &= 15 \text{ J} \end{aligned}

Hence, the efficiency of the electric motor will be:

\begin{aligned} \text{Efficiency} &= \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100 \% \\ &= \frac{50}{65} \times 100 \% \\ &= 76.9 \% \end{aligned}

Example 5

A crane is used to raise a weight of 200 N at a constant speed through a vertical height of 8.0 m in 4.0 s.

The efficiency of the crane is 20%. What is the electrical power needed to be supplied to the crane?

1. 2.0 kW
2. 20 kW
3. 4.0 kW
4. 5.0 kW

$$\text{Power output} = \frac{mgh}{t}$$

\begin{aligned} \text{Efficiency} &= \frac{\text{output}}{\text{input}} \\ \text{input} &= \frac{\text{output}}{\text{efficiency}} \\ &= \frac{200 \times \left(\frac{8}{4}\right)}{0.20} \\ &= 2000 \, \text{W} \end{aligned}

Example 6

a) i) Define power

ii) Hence, derive the equation P = Fv, where P is power, F is force and v is speed.

b) A motorized bicycle has a maximum of 0.23 hp, horse power. 1 hp = 746 W.

i) Along a horizontal road AB, the cyclist travels at a constant maximum speed of 33 km h-1. Calculate the total frictional force exerted on the system (bicycle and cyclist), if the efficiency of the motor is 60%.

ii) The cyclist then travels along a road BC inclined at 10° to the horizontal. Is it possible for the cyclist to move up the slope at the same constant speed of 33 km/h? Explain your answer.

iii) If the cyclist is to travel along the road inclined at 10° at the same constant speed of 33 km/h, calculate the new power required in hp(horse power) of the motor. The total frictional force acting on the system is now 0.8 times that in b)i), and the total mass of the system is 90 kg.

a) i) Power is defined as the rate of doing work.

ii) $$P = \frac{dW}{dt}$$

$$P = \frac{d(Fx)}{dt}$$

, where F is a constant force acting on a body and x is the displacement along the direction of the force.

Since force is constant,

\begin{aligned} P &= F \times \frac{dx}{dt} \\ P &= Fv \end{aligned}
P = Fv

b) i) Since $P = Fv$,

\begin{aligned} P &= Fv \\ \frac{60}{100} \times 0.23 \times 746 &= F \times \frac{33 \times 10^3}{3600} \\ F &= 11.231 \text{ N} \end{aligned}

Since speed is constant, the forward driving force F is equal to the total frictional force f acting on the system. Hence, $f = 11.2 \text{ N}$.

b) ii) No, it is not possible. This is because part of the power has to be used to do work against gravitational force.

b) iii)

\begin{aligned} P &= (f + m \cdot g \cdot \sin 10^\circ) \cdot v \\ P &= \left[(0.8 \times 11.231) + (90 \times 9.81 \times \sin 10^\circ)\right] \times \left(\frac{33 \times 10^3}{3600}\right) \\ P &= 1487.7 \, \text{W} \\ P &= \frac{1487.7}{746} \, \text{Hp} \\ P &= 2.0 \, \text{hp} \end{aligned}

Mini Physics

As the Administrator of Mini Physics, I possess a BSc. (Hons) in Physics. I am committed to ensuring the accuracy and quality of the content on this site. If you encounter any inaccuracies or have suggestions for enhancements, I encourage you to contact us. Your support and feedback are invaluable to us. If you appreciate the resources available on this site, kindly consider recommending Mini Physics to your friends. Together, we can foster a community passionate about Physics and continuous learning.

This site uses Akismet to reduce spam. Learn how your comment data is processed.