Power

Power

Applications of Power
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 =(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)
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 = \( P = \frac{{dm}} {{dt}} \cdot v^2 \)     
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
    \( P = \left( {R + ma} \right)V \)
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
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 = \( \frac{1} {2}.p_{inst} \, \);  
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 ‘’ 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.\( \frac{{dm}} {{dt}} \)    
    \( \frac{{F^| }} {F} = \frac{{V^| .\left( {\frac{{dm}} {{dt}}} \right)^| }} {{V\,.\,\frac{{dm}} {{dt}}}}\, = \,\frac{{(nV)\left( {n.\frac{{dm}} {{dt}}} \right)}} {{V\,\left( {\frac{{dm}} {{dt}}} \right)}}\, = \,n^2 \)
    \( \text{ }\boxed{\text{F}^\text{1} \text{ = n}^\text{2} \text{.F}} \)
    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.