Power - Meaning, Unit, Formula, FAQs

What is Power?

Power is defined as the rate of doing work or the rate of transfer of energy with respect to time. It indicates how fast work is performed by a body or a machine.

If an amount of work $W$ is done in time $t$, then the power $P$ is given by

$P=\frac{W}{t}$

The SI unit of power is the watt (W). One watt is defined as the power when one joule of work is done in one second.

$1 \mathrm{~W}=1 \mathrm{~J} \mathrm{~s}^{-1}$

Power is a scalar quantity.

Average Power

$
P_{a v g}=\frac{\Delta W}{\Delta t}=\frac{\int_0^t P \cdot d t}{\int_0^t d t}
$

$P_{\text {avg }}$ : Average power
$\Delta W$ : Change in work done
$\Delta t$ : 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{d W}{d t}=\vec{F} \cdot \vec{v}
$

• P: Instantaneous power
• $d w / d t$ : Rate of change of work
• $\vec{F}$ : Force vector
• $\vec{v}$ : Velocity vector

Difference Between Power and Energy

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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 $(\Delta K)$ :

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

$
\begin{aligned}
P & =\frac{\Delta K}{\Delta t}=\frac{300,000}{300} \\
P & =1000 \mathrm{~W}=1 \mathrm{~kW}
\end{aligned}
$
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 $F=m a$ and $a=\frac{d V}{d t}$ :

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

$
V \cdot \frac{d V}{d t}=\frac{C}{m}
$
Integrate:

$
\frac{V^2}{2}=\frac{C}{m} \cdot t \Longrightarrow V^2 \propto t \Longrightarrow V \propto t^{1 / 2}
$
Since $V=\frac{d x}{d t}$ :

$
\frac{d x}{d t} \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{d m}{d t} \cdot v^2
$
Substitute the given values:

$
\begin{gathered}
P=0.5 \cdot(5)^2 \\
P=0.5 \cdot 25=12.5 \mathrm{~W}
\end{gathered}
$
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{d m}{d t} \cdot v^2
$

Substituting the values:
- $\frac{d m}{d t}=0.5 \mathrm{~kg} / \mathrm{s}$
- $v=5 \mathrm{~m} / \mathrm{s}$

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

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