Informative line

Power

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?

A

Robin 

.

B

Alia 

C

Sara 

D

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. 

A

60 W

.

B

7 W

C

1 W

D

Zero 

Option C 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 Kf = Final kinetic energy

Ki = 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.)

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

where vi = 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

vf = 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 = Kf – Ki

= 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.  

A

– 8 W

.

B

7 W

C

5 W

D

– 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

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

  • As,

    \(dW = \vec{F}.\;{d\vec s}\)

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

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

   \(P\; = \vec{F}.\vec{v}\)

 

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.

image

\(v = u \;cos\;53° \)

\( =\;20 × \;\dfrac{3}{5}\)

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

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

\(P= F \,v\, cos\, 90° \)

\(P = 0\)

image

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

Option B 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.

A

12 W

.

B

10 W

C

14 W

D

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.

A

216 W

.

B

220 W

C

230 W

D

240 W

Option A is Correct

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