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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 forces 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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