Power - Meaning, Unit, Formula, FAQs

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

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

$P_{\text {avg }}$ : 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
- $\quad d W / d t$ : Rate of change of work
- $\vec{F}$ : Force vector
- $\vec{v}$ : 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:

$$
P=\frac{d K}{d t}
$$

- P: Power
- $d K / d t$ : 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 $(\Delta K)$ :

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