Power

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 ‘\(\rho\)’ is flowing through a pipe of cross section ‘A’ at speed ‘v’ , 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(AV\(\rho\))
As the pipe and liquid are same  \(\rho ^| \)= \(\rho\)    A|=A ,  \( V^| \)= nV
 a) Now as F =V. \( \frac{{dm}} {{dt}} \)   
    \( \frac{{F^| }} {F} = \frac{{V^| .\left( {\frac{{dm}} {{dt}}} \right)^| }} {{V\,.\,\frac{{dm}} {{dt}}}}\, = \,\frac{{(nV)\left( {n.\frac{{dm}} {{dt}}} \right)}} {{V\,\left( {\frac{{dm}} {{dt}}} \right)}}\, = \,n^2 \)  
       \( \text{ }\therefore \boxed{\text{F}^\text{1} \text{ = n}^\text{2} \text{.F}} \)
 
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 time