Power of a body is defined as the rate at which the body can do the work.
Average power \(({{P}_{\text{av}\text{.}}})=\frac{\Delta W}{\Delta t}=\frac{W}{t}\)
Instantaneous power \(({{P}_{\text{inst}\text{.}}})=\frac{dW}{dt}=\frac{\vec{F}.\,d\vec{s}}{dt}\,\,\, [As\,\, dW=\vec{F}.\,d\vec{s}]\)
\({{P}_{\text{inst}}}=\vec{F}.\,\vec{v}\) [As \(\vec{v}=\frac{d\vec{s}}{dt}\)]
i.e. power is equal to the scalar product of force with velocity.
Important Points
(1) Dimension : \([P]=[F]\,[v]=[ML{{T}^{-2}}]\,[L{{T}^{-1}}]\)
\(\therefore[P]=[M{{L}^{2}}{{T}^{-3}}]\)
(2) Units : Watt or Joule/sec [S.I.]
Erg/sec [C.G.S.]
Practical units : Kilowatt (KW), Mega watt (MW) and Horse power (hp)
Relations between different units :
\(1\,Watt=1\,Joule/\sec ={{10}^{7}}erg/\sec \)
\(1hp=746\,Watt\)
\(1\,MW={{10}^{6}}\,Watt\)
\(1\,KW={{10}^{3}}\,Watt\)
(3) If work done by the two bodies is same then power \(\propto \frac{1}{\text{time}}\)
i.e. the body which perform the given work in lesser time possess more power and vice-versa.
(4) As power = work/time, any unit of power multiplied by a unit of time gives unit of work (or energy) and not power, i.e. Kilowatt-hour or watt-day are units of work or energy.
\(1\,KWh={{10}^{3}}\frac{J}{sec}\times (60\times 60sec)=3.6\times {{10}^{6}}\,Joule\)
(5) The slope of work time curve gives the instantaneous power. As P = dW/dt = tanq
(6) Area under power-time curve gives the work done as \(P=\frac{dW}{dt}\)
\(\therefore W=\int{P\,dt}\)
\(\therefore\) W = Area under P-t curve
An automobile of mass m accelerates, starting from rest, while the engine supplies constant power P, its position and velocity changes w.r.t time.
(1) Velocity : As Fv = P = constant
i.e.\(m\frac{dv}{dt}v=P\) \(\left[ \text{As }F=\frac{mdv}{dt} \right]\)
or \(\int_{{}}^{{}}{v\,dv}=\int_{{}}^{{}}{\frac{P}{m}dt}\)
By integrating both sides we get \(\frac{{{v}^{2}}}{2}=\frac{P}{m}t+{{C}_{1}}\)
As initially the body is at rest i.e. v = 0 at t = 0, so \({{C}_{1}}=0\)
\(\therefore\ v={{\left( \frac{2Pt}{m} \right)}^{1/2}}\)
(2) Position : From the above expression \(v={{\left( \frac{2Pt}{m} \right)}^{1/2}}\)
or \(\frac{ds}{dt}={{\left( \frac{2Pt}{m} \right)}^{1/2}}\) \(\left[ \text{As }v=\frac{ds}{dt} \right]\)
i.e.\(\int_{{}}^{{}}{ds}=\int_{{}}^{{}}{{{\left( \frac{2Pt}{m} \right)}^{1/2}}\,dt}\)
By integrating both sides we get
\(s={{\left( \frac{2P}{m} \right)}^{1/2}}.\frac{2}{3}{{t}^{3/2}}+{{C}_{2}}\)
Now as at t = 0, s = 0, so \({{C}_{2}}=0\)
\(s={{\left( \frac{8P}{9m} \right)}^{1/2}}{{t}^{3/2}}\)
Power
Often it is interesting to know not only the work done on an object, but also the rate at which this work is done. We say a person is physically fit if he not only climbs four floors of a building but climbs them fast.Power is defined as the time rate at which work is done or energy is transferred.
The average power of a force is defined as the ratio of the work, W, to the total time t taken \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaBa % aaleaacaWGHbGaamODaaqabaGccqGH9aqpdaWcaaqaaiaadEfaaeaa % caWG0baaaaaa!3BCD! {P_{av}} = \frac{W}{t}\)
The instantaneous power is defined as the limiting value of the average power as time interval approaches zero,
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 % da9maalaaabaGaamizaiaadEfaaeaacaWGKbGaamiDaaaadaqadaqa % aiaaiAdacaGGUaGaaGOmaiaaigdaaiaawIcacaGLPaaaaaa!3FFA! P = \frac{{dW}}{{dt}}\left( {6.21} \right)\)
The work dW done by a force F for a displacement dr is dW = F.dr. The instantaneous power can also be expressed as
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGqb % Gaeyypa0JaamOraiaac6cadaWcaaqaaiaadsgacaWGYbaabaGaamiz % aiaadshaaaaabaGaeyypa0JaamOraiaac6cacaWG2bWaaeWaaeaaca % aI2aGaaiOlaiaaikdacaaIYaaacaGLOaGaayzkaaaaaaa!4518! \begin{array}{l} P = F.\frac{{dr}}{{dt}}\\ = F.v\left( {6.22} \right) \end{array}\)
where v is the instantaneous velocity when the force is F.
Power, like work and energy, is a scalar quantity. Its dimensions are [ML2T–3]. In the SI, its unit is called a watt (W). The watt is 1 J s–1. The unit of power is named after James Watt, one of the innovators of the steam engine in the eighteenth century.There is another unit of power, namely the horse-power (hp)
1 hp = 746 W
This unit is still used to describe the output of automobiles, motorbikes, etc.
We encounter the unit watt when we buy electrical goods such as bulbs, heaters and refrigerators. A 100 watt bulb which is on for 10 hours uses 1 kilowatt hour (kWh) of energy.
100 (watt) \(\times\) 10 (hour)
= 1000 watt hour
=1 kilowatt hour (kWh)
= 103 (W) 3600 (s)
= 3.6 \(\times\) 106 J
Our electricity bills carry the energy consumption in units of kWh. Note that kWh is a unit of energy and not of power.
EXAMPLE 11
An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 m s–1. The frictional force opposing the motion is 4000 N.Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.
ANSWER
The downward force on the elevator isF = m g + Ff = (1800 \(\times\) 10) + 4000 = 22000 N
The motor must supply enough power to balance this force. Hence,
P = F. v = 22000 \(\times\) 2 = 44000 W = 59 hp.