Power

Power

Power

Power is defined as the time rate at which work is done (or) rate at which energy is transferred is called as power.
If 'W' is the total work done by a force in a time interval 't', then the average power is
                 \( P_{av} = \frac{W} {t} = \frac{{\text{total work}}} {{\text{total time}}} \)
Instantaneous power is the dot product of force and velocity of body, provided the force does not change with time.
The instantaneous power is 
                     \( \begin{gathered} P_{inst} = \overrightarrow F .\overrightarrow V = FV\cos \theta \hfill \\ P_{inst} = \mathop {lt}\limits_{\Delta t \to 0} = \frac{{\Delta W}} {{\Delta t}} = \frac{{dW}} {{dt}};dW = \overrightarrow F .\overrightarrow {ds} \hfill \\ P_{inst} = \overrightarrow F .\frac{{\overrightarrow {ds} }} {{dt}};P_{inst} = \overrightarrow F .\overrightarrow V ;\left( {\frac{{\overrightarrow {ds} }} {{dt}} = \overrightarrow V } \right) \hfill \\ \to P_{inst} = FV\cos \theta \hfill \\ \end{gathered} \)
Where 'V' is the instantaneous velocity

Note : If the work done is constant then P \( \alpha \frac{1} {t} \). 

Power

DIMENSIONS AND UNITS OF POWER

(a) Power is a scalar quantity with dimensions \( [ML^2 T^{ - 3} ] \).
(b) S.I unit of power is J/sec (or) watt (W). CGS unit of power is erg/sec.
(c) Practical unit of power is horse power (Hp)
            1Hp=746W

Watt: The power of an agent is said to be one watt, if one joule of work is done in one second.
            1 watt = \( \frac{{\text{1 joule}}} {{1s}} = 1Js^{ - 1} \)

(d) The slope of W-t curve gives instantaneous power.
              

            \( P = \frac{{dW}} {{dt}} = \tan \theta \) 
            (as shown in fig.(a))

(e) The area under P-t graph gives workdone as  
                

              \( P = \frac{{dW}} {{dt}} \Rightarrow W = \int {P.dt} \)              

Power

RELATION BETWEEN AVERAGE POWER AND INSTANTANEOUS POWER

A particle starts from rest and moving with uniform  acceleration and gains velocity “V” in  it time ”t”.
    P_av = \( \frac{W} {t} \)= \( \frac{{\frac{1} {2}mV^2 }} {t} \) = ;   
      \( \text{P}_{\text{av}} \text{ = }\frac{\text{W}} {\text{t}}\text{ = }\frac{{\text{m\vec a} \cdot \vec V}} {\text{2}} = \frac{{\vec F \cdot \vec V}} {\text{2}} \)
we know instantaneous power is P_inst= \( \vec F.\vec V \) 
From the above it can be concluded that \( P_{av} = \frac{1} {2}P_{inst} \)   

Power

EFFICIENCY OF CRANE OR MOTOR 

The ratio of output power to the input power is called efficiency.
              \( \eta = \frac{{output\,power}} {{Input\,power}} \times \,100 \)     \( \eta = \,\,\frac{{p_0 }} {{p_i }} \times 100 \)
Efficiency can also be defined as the ratio of useful workdone to total energy spent.
* If \(\eta\) is the efficiency of motor, then  
            Total input power  \( P_i \, = \,\frac{{100}} {\eta } \times \frac{{mgh}} {t} \)

Power

POINTS TO BE REMEMBERED REGARDING POWER

1) When body moves with constant velocity,\( P_{inst} = P_{ave} \)  While if body starts from restand moves with constant acceleration, then \( P_{inst} = 2P_{ave} \)
2) In a straight line motion, if a constant force acts on a body, the power developed by this force varies linearly with time. Also power developed varies parabolically with displacement.
3)Power of a centripetal force in a circular motion is zero.

Power

APPLICATIONS OF POWER IN VARIOUS SITUATIONS

Application - 1
If a machine gun fires 'n' bullets per second such that mass of each bullet is 'm' and comming out with a velocity 'v' then the power of the machine gun is  
               \( P = \frac{{N\left( {\frac{1} {2}mV^2 } \right)}} {t} \)
    (where N bullets are fired in time 't' then n = N/t)
        P_av = \( \frac{1} {2}mnV^2 \)
    (power of heart=pressure x volume of blood pumped per second)

Application - 2
A conveyor belt moves horizontally  with a constant speed 'v' Gravel is falling on it at a rate of 'dm/dt' then,
    a) Extra force required to drive the belt is  \( F = \frac{{dm}} {{dt}} \cdot v \)
    b) Extra power required to drive the belt is
            P = FV =  \( P = \frac{{dm}} {{dt}} \cdot v^2 \)     

Application - 3
A car of mass 'm' is moving on a horizontal road with constant accelerction 'a'. If R is the resistantce offered to its motion, then the instaneous power of the engine when its velocity is  'v'
    Net force on the car is F – R = ma
    driving force of the engine is F = R + ma
    Instantaneous power     P = F. V
               \( P = \left( {R + ma} \right)V \)

Power

Application - 4
The car moves on a rough horizontal road with a constant speed 'V' then the instaneous power of engine is
    P = F.V (V constant)  
    But F = f 
    P = fV (Here f = frictional force on rough  horizontal surface)  
    P =\(\mu\) mg. V

Application - 5
A body of mass 'm' is initially at rest. By the application of constant force its velocity changes to "V₀" in time ' t_o' then
             \( \begin{gathered} v = u + at \hfill \\ v_0 = at_0 \hfill \\ \end{gathered} \)
    acceleration of the body is  \( a\, = \,\frac{{v_0 }} {{t_0 }} \)
a) Find instantaneous power at an instant of time 't' is 
        P = F.V = (ma) (at) = m a² t 
           \( P_{\operatorname{in} st} = m\left( {\frac{{v_0 }} {{t_0 }}} \right)^2 t \)                         
b)  Average power during the time 't' is
        P_av =\( \frac{1} {2}.p_{inst} \, \) ;  

Power

Application - 6

A  motor pump is used to deliver water at a certain rate from a given pipe. To obtain 'n' times water from the same pipe in the same time by what amout (a) force and (b) power of the motor should be increased.
If a liquid of density ‘\(\rho\)’ is flowing through a pipe of cross section ‘A’ at speed , the mass comming out per second will be.
           \( \frac{{dm}} {{dt}} = AV\rho \)      
To set ‘n’ times water in the same time
        \( \left( {\frac{{dm}} {{dt}}} \right)^1 = n\left( {\frac{{dm}} {{dt}}} \right) \)
    A^|V^| \(\rho^|\)= n(A\(V\rho\))
    As the pipe and liquid are same  \( \rho ^| =\rho \)     
    A^|=A ,   V^|= nV
    a)  Now as F = \( v.\frac{{dm}} {{dt}} \).    
         \( \frac{{F^| }} {F} = \frac{{V^| .\left( {\frac{{dm}} {{dt}}} \right)^| }} {{V\,.\,\frac{{dm}} {{dt}}}}\, = \,\frac{{(nV)\left( {n.\frac{{dm}} {{dt}}} \right)}} {{V\,\left( {\frac{{dm}} {{dt}}} \right)}}\, = \,n^2 \)
             \( \therefore\text{ }\boxed{\text{F}^\text{1} \text{ = n}^\text{2} \text{.F}} \)
    b)  and as P = F.
          \( \frac{{P^| }} {P} = \frac{{F^| V^| }} {{FV}} = \frac{{(n^2 F)(n\,V)}} {{F\,V}} = n^3 \)     \( \therefore \boxed{\text{P}^\text{1} \text{ = n}^\text{3} \text{.P}} \)
To get ‘n’ time of water. force must be increased n² times while power n³ times.

Application - 7
If pump lifts water from depth h and delivers at a rate \( \frac{{dm}} {{dt}} \) with a velocity v then
power deliverd,P= \( \frac {dm} {dt}\left[ {gh + \frac{{v^2 }} {2}} \right] \)

Power

Application- 8
A car of mass 'm' is moving on a horizontal road with constant accelerction 'a'. If R is the resistance offered to its motion, then the instaneous power of the engine when its velocity 'V' is given by
        Net force on the car is F - R = ma
        driving force of the engine is F = R + ma
                 
  
Instantaneous power P = F. V
            P = (R+ma)V

Application -9
An automobile of mass 'm' acceleretes, starting from rest, while the engine supplies constant power, its position and velocity change w.r.t time

a)Velocity: As FV = P = constant
i.e.,  \( \frac{{dV}} {{dt}}.V = P \)           (F=m.\( \frac{{dV}} {{dt}} \))
or      \( \frac{{dV}} {{dt}}.\int V dV = \int {\frac{p} {m}.dt} \)       
on integrating both sides we get  \( \frac{{V^2 }} {2} = \frac{p} {m}.t + c_1 \)
As initialy the body in at rest  
    i. e V=0 at t=0 \( \Rightarrow c_1 = 0 \)
         \( V = \left[ {\frac{{2pt}} {m}} \right]^{\frac{1} {2}} \)         \( \therefore V\alpha t^{\frac{1} {2}} \)

b)Position: From the above expression
         \( V = \left[ {\frac{{2pt}} {m}} \right]^{\frac{1} {2}} \)  (or)  \( \frac{{ds}} {{dt}} = \left[ {\frac{{2pt}} {m}} \right]^{\frac{1} {2}} \)  (As V=\( \frac{{ds}} {{dt}} \))
              \( \therefore \int {ds} = \int {\left[ {\frac{{2pt}} {m}} \right]^{\frac{1} {2}} dt} \)
By integrating both sides we get
        s= \( \left[ {\frac{{2pt}} {m}} \right]^{\frac{1} {2}} \)  \( \left[ {\frac{{2pt}} {m}} \right]^{\frac{1} {2}} .\frac{2} {3}.t^{\frac{3} {2}} + c_2 \)
Now as at t=0, s = 0  \( \Rightarrow c_2 = 0 \)
        s= \( \left[ {\frac{{8p}} {{9m}}} \right]^{\frac{1} {2}} .t^{\frac{3} {2}} \)    \( \therefore s\alpha t^{\frac{3} {2}} \)