Power

Power is the time rate at which work is done or energy is transferred.
For example, a robot A can run 10 km and it finishes every km in 10 minutes.
Another robot B can run only 5 km but it finishes every km in 3 minutes.
It implies that robot A has more energy but less power.

Illustration Questions

In a competition, Robin, Sara and Alia climb a rope in 2 min, 3 min and 6 min respectively. Who has more power?

A Robin

B Alia

C Sara

D All have same power

As all climb equal length of rope, work done by them is same.

But Robin takes minimum time to climb the rope.

\(\therefore \) She has more power than others.

In a competition, Robin, Sara and Alia climb a rope in 2 min, 3 min and 6 min respectively. Who has more power?

Robin

Alia

Sara

All have same power

Option A is Correct

Average Power

If an external force is applied to an object, the work done by this force in time interval \(\Delta\;t\) is W.
The average power during this time interval is

\(P_{avg}\; = \dfrac{W}{\Delta\;t}\)

Unit of average power is Joules per second or watt (W)

Illustration Questions

Ana can do 60 J of work in 1 minute, calculate the average power of Ana.

A 60 W

B 7 W

C 1 W

D Zero

Amount of work done = W = 60 J

time interval, \(\Delta\,t\) = 1 minute = 60 second

Avg power, \(P_{avg} = \dfrac{W}{\Delta\;t }\)

\(\dfrac{60}{60}\) = 1 W

Ana can do 60 J of work in 1 minute, calculate the average power of Ana.

60 W

7 W

1 W

Zero

Option C is Correct

Instantaneous Power When Work is Given as a Function of Time

W = f (t)

As \(P\;=\;\dfrac{dW}{dt}\)

\(P = \dfrac{d}{dt}\;[f(t)]\)

For instantaneous power at time t, put the value of t after the differentiation of work.

Illustration Questions

If work done by a force is given as W (t) = 3 t2 + 2 t, find the power delivered by force at t = 2 second.

A 12 W

B 10 W

C 14 W

D 16 W

Work W(t) = 3t² + 2t

\(P\;=\; \dfrac{dW(t)}{dt}\)

\(P= \dfrac{d}{dt}\;(3t^2\;+\;2t)\)

\(P = 6 t + 2\)

Power at t = 2 sec

\(P = (6× 2)\;+ 2\)

\(P= 14\, W\)

If work done by a force is given as W (t) = 3 t2 + 2 t, find the power delivered by force at t = 2 second.

12 W

10 W

14 W

16 W

Option C is Correct

Calculation of Power when Position x is given as a Function of Time

If position \(x\) is given \(x(t)\)
As \(v = \dfrac{dx}{dt}\;\;and \;\;a = \dfrac{d^2x}{dt^2}\)
Force, F = m a

\(F = m × \dfrac{d^2x(t)}{dt^2}\)

Power P = F.v

\(P = \dfrac{md^2x(t)}{dt^2}\;× \;\dfrac{dx(t)}{dt}\)

Illustration Questions

A particle of mass 2 kg is moving under the influence of a force. If its position is given as \(x(t) = 4 t^3 -3t^2.\) Calculate the power delivered by this force at time t = 1 second.

A 216 W

B 220 W

C 230 W

D 240 W

Position \(x\)= 4 t³ – 3 t²

Velocity , \(v = \dfrac{dx}{dt}\)

\(=\;\dfrac{d}{dt}\;\left(4t^3\;-\;3t^2\right)\)

= 12 t² – 6 t

Acceleration, \(a = \dfrac{dv}{dt}\)

\(=\;\dfrac{d}{dt}\;\left(12t^2\;-\;6t\right)\)

=24 t – 6

Power, P = \(\vec{F}.\vec{v}\)

= m a. v

\( = 2 \;(24 t \;-\;6)\;× \;(12 t^2 - 6 t)\)

Power at t = 1 second

= \(2 × (24 - 6 )× (12-6)\)

= 2 × 18 × 6

= 216 W

A particle of mass 2 kg is moving under the influence of a force. If its position is given as \(x(t) = 4 t^3 -3t^2.\) Calculate the power delivered by this force at time t = 1 second.

216 W

220 W

230 W

240 W

Option A is Correct

Calculation of Average Power if Change in Kinetic Energy is Known

According to work-energy theorem

\(W = \;\Delta \;KE\)

\(W = (K_f-K_i)\)

\(P_{avg} = \dfrac{W}{\Delta t}\)

\(P_{avg} = \dfrac{K_f - K_i}{\Delta t}\)

where K_f = Final kinetic energy

K_i = Initial kinetic energy

\(\Delta\,t\) = Time interval

Illustration Questions

A particle of mass 2 kg, is thrown vertically upwards with an initial velocity of 3 m/s. If particle starts journey at t = 0 and reaches the maximum height in 3 sec, find the average power during this time interval.

A – 8 W

B 7 W

C 5 W

D – 3 W

Calculation of initial kinetic energy (K.E.)_i

\((K.E.)_i\; =\; \dfrac{1}{2} mv_i^2\)

where v_i = Initial velocity

\(=\dfrac{1}{2}\; × 2\; × (3)^2\)

\(=9 \,J\)

Calculation of final kinetic energy (K.E)_f

At maximum height, the velocity of particle becomes zero

v_f = 0

(K.E.)_f = \(\dfrac{1}{2} m\;v_f^2\)

\(=\dfrac{1}{2} × m × 0\)

= 0

From work energy theorem

W = \(\Delta\) K.E

W = K_f – K_i

= 0 – 9

W = – 9 J

Average power,

\(P_{avg}\; = \; \dfrac{W}{\Delta t}\)

= \(\dfrac{-9}{3}\)

\(P_{avg} = -3\;W\)

Here, negative sign implies, the particle is losing its power.

A particle of mass 2 kg, is thrown vertically upwards with an initial velocity of 3 m/s. If particle starts journey at t = 0 and reaches the maximum height in 3 sec, find the average power during this time interval.

– 8 W

7 W

5 W

– 3 W

Option D is Correct

Instantaneous Power

Power at given instant of time is known as instantaneous power.
Instantaneous power is the limiting value of the average power as

\(\Delta t \) → 0

\(P=\lim\limits_{\Delta t\rightarrow0}\; \dfrac{W}{\Delta t\;}\;\;=\;\;\dfrac{dW}{dt}\)

\(dW = \vec{F}.\;{d\vec s}\)

\(\dfrac{dW}{dt}\; =\;\vec{F}.\;\dfrac{d\vec{s}}{dt}\)

\(\dfrac{dW}{dt}\;=\;\vec{F}. \vec{v}\)

\(P\; = \vec{F}.\vec{v}\)

Illustration Questions

A particle of mass 2 kg is projected making an angle 53 ° with the horizontal at a speed of 20 m/sec. Find the power delivered by the force of gravity at the highest point of trajectory. \((cos\,53 °=\dfrac{3}{5})\)

A – 320 W

B 0

C 100 W

D 10 W

Force of gravity

F = mg

= 2 × 10 = 20 N

At the highest point, the angle between velocity and mg becomes 90°, as shown in figure.

\(v = u \;cos\;53° \)

\( =\;20 × \;\dfrac{3}{5}\)

\(v\;=\;12\; m/s\)

\(P = \vec{F}.\vec{v}\)

\(P= F \,v\, cos\, 90° \)

\(P = 0\)

A particle of mass 2 kg is projected making an angle 53 ° with the horizontal at a speed of 20 m/sec. Find the power delivered by the force of gravity at the highest point of trajectory. \((cos\,53 °=\dfrac{3}{5})\)

– 320 W

100 W

10 W

Option B is Correct