Power - Meaning, Unit, Formula, FAQs

Power in physics refers to the amount of work done or energy transferred per unit of time. It is expressed in watts (W) and a watt is the equivalent of one joule per second. Power measures, how fast the energy is used or converted to another form. In electrical systems, for e.g. power = voltage x current (in a charge-neutral case, what this really tells you is how much electrical energy is being consumed, or being pumped, per unit of time. Power is important in running the various types of equipment and machines we use in everyday life. Generally, the higher the power rating, the more work a device can do in less time.

In this article, we are going to read about power and different types of power and also see some solved examples, which belong to the chapter work, energy, and Power, which is one of the important chapters in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), more than fifteen questions have been asked on this concept. And for NEET almost six questions were asked from this concept.

Let's read this entire article to gain an in-depth understanding of the concept of power.

Define Power

Power is defined as the rate at which work is done or energy is transferred.

$M{L}^2{T}^{-3}$

• $M$: Stands for mass
• $L$: Stands for length
• $T$: Stands for time

Units - Watt or Joule/sec (in SI), Erg/sec (in CGS)

Average power

$$P_{avg}=\frac{\mathrm{\Delta }W}{\mathrm{\Delta }t}=\frac{{\int }_0^{t}P\cdot dt}{{\int }_0^{t}dt}$$

$P_{\text{avg }}$ : Average power
$\mathrm{\Delta }W$ : Change in work done
$\mathrm{\Delta }t$ : Change in time
$P$: Power at a given time

The average power is computed over a time interval by taking the total work done divided by the total time taken.

Instantaneous Power:

$$P=\frac{dW}{dt}=\overrightarrow{F}\cdot \overrightarrow{v}$$

- P: Instantaneous power
- $dW/dt$ : Rate of change of work
- $\overrightarrow{F}$ : Force vector
- $\overrightarrow{v}$ : Velocity vector

Instantaneous power is the dot product of the force applied and the velocity of the object. This represents the power being delivered at any specific moment in time.

Power and Kinetic Energy:

$$P=\frac{dK}{dt}$$

- P: Power
- $dK/dt$ : Rate of change of kinetic energy

Power can also be defined as the rate of change of kinetic energy, connecting work-energy principles to power.

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Solved Example Based on Power

Example 1: An engine of a car of mass m = 1000 Kg changes its velocity from 5 m/s to 25 m/s in 5 minutes. The power (in KW) of the engine is

1) 5

2) 2

3) 1

4) 4

Solution:

Calculate the change in kinetic energy $(\mathrm{\Delta }K)$ :

$$\begin{array}{c}\mathrm{\Delta }K=\frac{1}{2}\cdot 1000\cdot ((25{)}^2-(5{)}^2)\\ \mathrm{\Delta }K=\frac{1}{2}\cdot 1000\cdot (625-25)\\ \mathrm{\Delta }K=\frac{1}{2}\cdot 1000\cdot 600=300,000\mathrm{J}\end{array}$$
Now, calculate the power:

$$\begin{array}{rl}P& =\frac{\mathrm{\Delta }K}{\mathrm{\Delta }t}=\frac{300,000}{300}\\ P& =1000\mathrm{W}=1\mathrm{kW}\end{array}$$
Final Answer: Hence, the power of the engine is 1 kW.
Correct Option: (3).

Example 2: A constant power-delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to :

1) t2/3
2) t
3) t3/2
4) t1/2

Solution:

Power delivered by the machine is constant:

$$P=F\cdot V$$
Substitute $F=ma$ and $a=\frac{dV}{dt}$ :

$$P=m\cdot V\cdot \frac{dV}{dt}$$
Given $P=C$ (constant):

$$V\cdot \frac{dV}{dt}=\frac{C}{m}$$
Integrate:

$$\frac{V^2}{2}=\frac{C}{m}\cdot t⟹V^2\propto t⟹V\propto t^{1/2}$$
Since $V=\frac{dx}{dt}$ :

$$\frac{dx}{dt}\propto t^{1/2}$$

Integrate again:

$$x\propto t^{3/2}$$
Final Answer:
Correct Option: (3).

Example 3: Sand is being dropped from a stationary dropper at a rate of 0.5 kg−1 on a conveyor belt moving with a velocity of 5 ms−1. The power needed to keep the belt moving with the same velocity will be :

1) 1.25 W
2) 2.5 W
3) 6.25 W
4) 12.5 W

Solution:

The power needed to maintain the belt's motion is:

$$P=\frac{dm}{dt}\cdot v^2$$
Substitute the given values:

$$\begin{array}{c}P=0.5\cdot (5{)}^2\\ P=0.5\cdot 25=12.5\mathrm{W}\end{array}$$
Final Answer:
The power required is 12.5 W . Correct Option: (4).

Example 4: Sand is being dropped from a stationary dropper at a rate of $0.5\mathrm{kg}-1$ on a conveyor belt moving with a velocity of $5\mathrm{ms}-1$. The power needed to keep the belt moving with the same velocity will be:

1) 1.25 W
2) 2.5 W
3) 6.25 W
4) 12.5 W

Solution:

The power needed to keep the conveyor belt moving with the same velocity is given by:

$$P=\frac{dm}{dt}\cdot v^2$$

Substituting the values:
- $\frac{dm}{dt}=0.5\mathrm{kg}/\mathrm{s}$
- $v=5\mathrm{m}/\mathrm{s}$

$$\begin{array}{c}P=(0.5)\cdot (5{)}^2\\ P=0.5\cdot 25=12.5\mathrm{W}\end{array}$$

Final Answer:
The power required is 12.5 W .
Correct Option: (4).