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Power Definition, Formula ,Calculation ,Examples




(12) Power

  • Power is defined as rate of doing the work
  • if ΔW amount of work is done in time interval Δt ,the instantaneous power delivered will be
    $P=\frac {\Delta W}{ \Delta t}$
    or
    $P=\frac {dW}{dt}$
  • For total work done W in total time t,then average power is defined
    $P_{avg}=\frac {W}{t}$
  • If Power (P) does not vary with time ,then $P=P_{avg}$
  • the instantaneous power can also be written as
    $P=\frac {dW}{dt}$
    Now as dW=Fdx
    $P=\frac {F dx}{dt}$
    $P = F \frac {dx}{dt}$
    Now velocity is given as $v= \frac {dx}{dt}$
    Therefore,
    $P =F.v$
    This can be written in vector form as
    P=F.v
    This is scalar dot product of force and velocity vector.
    If the Force and velocity vector are at angle,then
    $P= Fvcos \theta$
  • SI unit of power is joule/sec also called watt
  • Power, like work and energy, is a scalar quantity. Its dimensions are $[ML^2T^{-3}]$.
  • Another unit of power is horsepower(hp)
    1Hp=746 watt




Example-1
An elevator can carry a maximum load of 1000 kg (elevator + passengers) is moving up with a constant speed of 2 m/s. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts ?
Solution
The downward force on the elevator is
$F = m g + f_{fric}$
$f = (1000 \times 10) + 4000 = 14000 N$
The motor must supply enough power to balance this force. Hence,
$P = F. v = 14000 \times 2 = 28000 W$

Example-2
A truck of mass 1000 kg accelerates on the level road from 10 m/s to 20 m/s in 5 sec. The frictional force present is 4000N. Find the average power delivered by the engine.?
Solution
Acceleration of truck = $\frac {20 -10}{5} = 2 m/s^2$
So total force delivered by the engine = $ma + f_{fric} = 1000 \times 2 + 4000 = 6000 N$
Average power = F X average velocity = 6000 X 30 = 18000W


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