Instantaneous Power & Average Power

Power :  Power is defined as the rate of work done.

$ \displaystyle P = \frac{dW}{dt} $

This is called Instantaneous Power .

$ \displaystyle dW = \vec{F}.\vec{dx} $

$ \displaystyle P_{ins} = \frac{\vec{F}.\vec{dx}}{dt} $

$ \displaystyle P_{ins} = \vec{F}.\vec{v} $

If the force is variable, we calculate the average power as

$ \displaystyle P_{avg} = \frac{\int_{0}^{t} P dt}{\int_{0}^{t} dt} $

$ \displaystyle Average Power = \frac{work \; done}{time} $

$ \displaystyle P_{avg} = \frac{W}{t}= \frac{m v^2}{2 t} $

$ \displaystyle P_{avg} = \frac{1}{2}mv\frac{v}{t} = \frac{1}{2}\vec{F}.\vec{v}$

$ \displaystyle P_{avg} = \frac{1}{2}P_{ins} $

Power can also be expressed as the rate of change of kinetic energy.

Let a body of mass m move with a velocity v. The kinetic energy of the body is

$ \displaystyle K = \frac{1}{2}m v^2 $

$ \displaystyle \frac{dk}{dt} = \frac{1}{2}\frac{d}{dt}(m v^2) $

$ \displaystyle = m v. \frac{dv}{dt} $

$ \displaystyle = m \frac{dv}{dt} .v $

= Fext . v

= P

Therefore, $ \displaystyle P = \frac{dK}{dt} $

Solved Example : A particle is projected with a speed v at an angle θ with the horizontal. Find the mean power delivered by gravity during the ascent of the particle.

Solution : The magnitude of mean power for

$ \displaystyle P_m = \frac{1}{t_0} \int_{0}^{t_0} \vec{m g}.\vec{v’_y} dt $

$ \displaystyle P_m = \frac{1}{t_0} \int_{0}^{t_0} (-m g v’_y) dt $

Where v’y = vy – gt

$ \displaystyle P_m = \frac{1}{t_0} \int_{0}^{t_0} mg (v_y – gt ) dt \quad (numerically) $

Since vy = t0/g

$ \displaystyle P_m = \frac{m g v_y}{2} = \frac{m g vsin\theta}{2} $

Exercise : Two bodies of masses m1 and m2 (m2 > m1) are connected by a light inextensible string which passes through a smooth fixed pulley. What is the instantaneous power delivered by an external agent to pull m1 with constant (a) velocity v (b) acceleration a at any instant t

Exercise : A small body of mass m is located on a horizontal plane at the point O. The body acquires a horizontal velocity vo . Find the mean power developed by the friction force during the motion, if the coefficient of friction μ = 0.27, m = 1.0 kg and vo = 1.5 m/s

Also Read :

Work done by a Constant Force & Variable Force
Conservative & Non-conservative Forces
Work done by Spring Force
Kinetic Energy , Work Energy Theorem
Conservation of Energy & Significance of potential energy

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