Power

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Power of a body is defined as the rate at which the body can do work, i.e.,

$ \displaystyle \begin{array}{l}\text{Power}=\text{ Rate of doing work}\\\therefore \text{ Power =}\dfrac{{\text{Work}}}{{\text{Time}}}\ldots (1)\end{array}$

The power of an agent indicates how quickly it can perform work. When a body takes less time to do a certain amount of work, its power is considered to be greater, and vice-versa.

The power of an agent can also be expressed in terms of the force applied $ \vec{F} $ and the velocity $ \vec{v} $ of the body:

\[
\text{As } P = \frac{\text{Work}}{\text{Time}} = \frac{W}{t} = \frac{\vec{F} \cdot \vec{s}}{t}
\]

\[
P = \vec{F} \cdot \vec{v} \ldots (2)
\]

Where $ \vec{v} = \dfrac{\vec{s}}{t} $

Thus, power can be expressed as the dot product of applied force and velocity of the body.

If $ \theta $ is the angle between $ \vec{F} $ and $ \vec{v} $, then

\[
P = \vec{F} \cdot \vec{v} = Fv \cos \theta
\]

However, when $ \vec{v} $ is along $ \vec{F} $, $ \theta = 0^\circ $, so

$ \displaystyle \therefore P=Fv\cos {{0}^{{}^\circ }}=Fv$

Since power is the ratio of two scalar quantities $ W $ and $ t $, it is considered a scalar quantity. Hence, it is expressed as the dot product of $ F $ and $ v $.

The dimensions of power can be derived as follows:

\[
P = \frac{W}{t} = \frac{M^1L^2T^{-2}}{T} = M^1L^2T^{-3}
\]

Units of power:
In the SI system, the fundamental unit of power is the watt, symbolized by $ W $.

Using the formula

\[
P = \frac{W}{t}
\]

\[
1 \text{ watt} = \frac{1 \text{ joule}}{1 \text{ second}}
\]

which implies

\[
1W = 1J \cdot s^{-1}
\]

Therefore, power of an agent is said to be one watt, it it can do one joule of work in one second.

Bigger units of power

$ \displaystyle \begin{array}{l}1\text{ kilo watt }=\text{ 1000 watt}\\\Rightarrow 1\text{ kW }={{10}^{3}}\text{ W}\\\text{and }\\\text{1 mega watt }=1,000,000\text{ watt}\\\Rightarrow 1\text{ MW}={{10}^{6}}\text{ W}\end{array}$

In the c.g.s. system, the unit of power is $ 1 \text{ erg} \cdot s^{-1} $.

Thus,

\[
1W = 1J \cdot s^{-1} = 10^7 \text{ erg} \cdot s^{-1}
\]

In the SI system, the gravitational unit of power is expressed as $ \text{kg} \cdot f \cdot \text{m} \cdot s^{-1} $, while in the c.g.s. system, it is $ \text{g} \cdot f \cdot \text{cm} \cdot s^{-1} $.

Another commonly used unit of power, especially in engineering contexts, is horsepower (h.p.).

\[
1 \text{ h.p.} = 746W
\]

The instantaneous power is indicated by the slope of the work versus time graph:

\[
P = \frac{dW}{dt} = \tan \theta
\]

This relationship is illustrated in figure below.

Additionally, the area under the power-time graph represents the work done, as:

\[
P = \frac{dW}{dt}
\]

which indicates

$ \displaystyle W=\int{P}dt$ $ \displaystyle =\text{area under the }P-t\text{ curve}$

This is depicted in below.

Example 1:

A lift with a mass of $ 500 \, \text{kg} $ is to be raised at a uniform velocity of $ 0.4 \, \text{m/s} $. What is the minimum horsepower required for the motor?

Solution: Given $ m = 500 \, \text{kg} $, $ v = 0.4 \, \text{m/s} $, and $ P = ? $

\[
P = F \cdot v = (mg) \cdot v
\]

\[
= 500 \times 9.8 \times 0.4 = 1960 \, \text{watts}.
\]

Assuming no frictional losses in the motor, the minimum horsepower required $ P $ is:

\[
\frac{1960}{746} \text{ h.p.} \approx 2.62 \text{ h.p.}
\]

Example 2:

A tube-well pumps out 2400 kg of water per minute. If water is coming out with a velocity of $ 3 \, \text{ms}^{-1} $, what is the power of the pump? How much work is done, if the pump runs for 10 hours?

Solution:

Mass of water pumped per second, $ M = \dfrac{2400}{60} = 40 \, \text{kg} $

Velocity of water, $ v = 3 \, \text{ms}^{-1} $

Kinetic energy of water coming out per second $ = \dfrac{1}{2} M v^2 = \dfrac{1}{2} \times 40 \times (3)^2 = 180 \, \text{J} $

Therefore, power of the pump $ P = 180 \, \text{Js}^{-1} = 180 \, \text{watt} $

Work done $ \displaystyle =\text{power}\times \text{time}$

$ \displaystyle =180\times 10=1800\text{watt hour}$

$ \displaystyle \begin{array}{l}=1800\times 3600=6480000\text{joule}\\=6.48\times {{10}^{6}}\text{joule}\end{array}$

Example 3:

A man cycles up a hill, whose slope is $ 1 $ in $ 20 $ with a velocity of $ 6.4 \, \text{km h}^{-1} $ along the hill. The weight of the man and the cycle is $ 98 \, \text{kg} $. What work per minute is he doing? What is his horse power?

Solution:

As shown in figure, if the inclination of the hill with the horizontal is $ \theta $,

then:

\[
\sin \theta = \dfrac{1}{20}
\]

(The slope of $ 1 $ in $ 20 $ implies that if $ BC = 1 $, then $ AB = 20 $)

The weight $ W $ of the man and cycle can be resolved into two components. The component along the hill is $ W \sin \theta $. In order to cycle up, the force equal to $ W \sin \theta $ has to be applied in the upward direction as shown in the figure.

Therefore, force applied, $ F = W \sin \theta $

\[
= 98 \times \dfrac{1}{20} = 4.9 \, \text{kgf} = 4.9 \times 9.8 \, \text{N}
\]

Velocity of the man, $ v = 6.4 \, \text{km h}^{-1} $

\[
= 6.4 \times \dfrac{1000 \times 1000}{3600} \, \text{ms}^{-1} = \dfrac{16}{9} \, \text{ms}^{-1}
\]

Power of man, $ \displaystyle P=Fv=4.9\times 9.8\times \frac{{16}}{9}\text{W}$

$ \displaystyle =86.4\text{W}=0.1144\text{h}\text{.p}\text{.}$

Work done in 1 minute

$ \displaystyle =4.9\times 9.8\times \dfrac{{16}}{9}\times 60\text{J}=5122.13\text{J}$