Informative line

# 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.

#### 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)

#### 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

#### 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,

$$dW = \vec{F}.\;{d\vec s}$$

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

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

$$P\; = \vec{F}.\vec{v}$$

#### 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. $$(cos\,53 °=\dfrac{3}{5})$$

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.

#### 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

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

#### 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