Summary:
This is Lecture 6 for the topic “Work, Energy and Power”. You can find all resources related with Work, Energy and Power by clicking on this link
1. The Definition of Power
While Work measures how much energy is transferred, Power measures how fast that energy is transferred. Power is defined as the time rate of doing work or transferring energy.
- Type: Scalar quantity.
- SI Unit: Watt (W), where $1 \text{ W} = 1 \text{ J/s}$.
- Dimensional Formula: $[\text{M L}^2 \text{T}^{-3}]$
- Practical Units: Horsepower (hp), where $1 \text{ hp} \approx 746 \text{ W}$.
2. Average Power ($P_{avg}$)
Average power is the total work done divided by the total time taken. It provides a macroscopic view of the energy transfer rate over an interval $\Delta t$.
$$P_{avg} = \frac{\Delta W}{\Delta t} = \frac{\Delta E}{\Delta t}$$
3. Instantaneous Power ($P_{inst}$)
Instantaneous power is the power delivered at a specific, infinitesimally small instant of time. We use calculus to define this as the time derivative of work:
$$P_{inst} = \frac{dW}{dt}$$
Since an infinitesimally small amount of work is $dW = \vec{F} \cdot d\vec{r}$, we can substitute this into the power equation:
$$P_{inst} = \frac{\vec{F} \cdot d\vec{r}}{dt} = \vec{F} \cdot \left( \frac{d\vec{r}}{dt} \right)$$
Because velocity $\vec{v} = \frac{d\vec{r}}{dt}$, we get the most important formula for power in dynamics:
$$P_{inst} = \vec{F} \cdot \vec{v}$$
- Instantaneous power is the dot product of the force vector and the velocity vector.
- $P_{inst} = Fv \cos \theta$, where $\theta$ is the angle between the force and the velocity.
4. Graphical Interpretation of Power
Just as the area under a Force-Displacement ($F-x$) graph gives Work, the area under a Power-Time ($P-t$) graph gives the total work done (or total energy transferred) during that time interval.
$$\Delta W = \int P dt = \text{Area under } P-t \text{ curve}$$
5. Special JEE-Mains Case: Constant Power Kinematics
A highly tested concept in JEE-Mains is a vehicle (or particle) accelerating from rest under a constant power $P_0$. How do its velocity and displacement depend on time?
Starting with $P_0 = Fv$ and knowing $F = ma = m \frac{dv}{dt}$:
$$P_0 = \left(m \frac{dv}{dt}\right) v$$
$$v dv = \frac{P_0}{m} dt$$
Integrating both sides from rest ($t=0, v=0$):
$$\int_0^v v dv = \frac{P_0}{m} \int_0^t dt$$
$$\frac{v^2}{2} = \frac{P_0}{m} t \implies v = \sqrt{\frac{2P_0}{m}} t^{1/2}$$
Key Takeaway 1: Velocity is proportional to the square root of time ($v \propto t^{1/2}$).
To find displacement $x$, substitute $v = \frac{dx}{dt}$:
$$\frac{dx}{dt} = \sqrt{\frac{2P_0}{m}} t^{1/2}$$
$$\int_0^x dx = \sqrt{\frac{2P_0}{m}} \int_0^t t^{1/2} dt$$
$$x = \sqrt{\frac{2P_0}{m}} \left( \frac{2}{3} t^{3/2} \right)$$
Key Takeaway 2: Displacement is proportional to time to the power of 1.5 ($x \propto t^{3/2}$).
6. Pumping Water (Mass Flow Rate)
Another classic problem involves a pump lifting water. If a pump lifts a mass $m$ of water to a height $h$ and ejects it with velocity $v$ in time $t$:
$$W_{total} = \text{Change in Potential Energy} + \text{Change in Kinetic Energy}$$
$$W = mgh + \frac{1}{2}mv^2$$
$$P_{avg} = \frac{W}{t} = \left(\frac{m}{t}\right)gh + \frac{1}{2}\left(\frac{m}{t}\right)v^2$$
Here, $\left(\frac{m}{t}\right)$ is the mass flow rate (e.g., kg/s). If the water density is $\rho$, and the volume flow rate is $Q$ (e.g., $m^3/s$), then mass flow rate is $\rho Q$.
JEE-Mains/NEET Practice Questions
Question 1: Vector Power Formulation
A body of mass 2 kg is initially at rest. A constant force $\vec{F} = (3\hat{i} + 4\hat{j})$ N acts on it. What is the instantaneous power delivered by the force at $t = 4$ seconds?
(a) 50 W
(b) 100 W
(c) 75 W
(d) 25 W
Solution: (a)
First, find acceleration: $\vec{a} = \frac{\vec{F}}{m} = \frac{3\hat{i} + 4\hat{j}}{2} = 1.5\hat{i} + 2.0\hat{j}$ m/s$^2$.
Next, find velocity at $t = 4$ s using $\vec{v} = \vec{u} + \vec{a}t$:
$\vec{v} = 0 + (1.5\hat{i} + 2.0\hat{j})(4) = 6\hat{i} + 8\hat{j}$ m/s.
Instantaneous Power $P = \vec{F} \cdot \vec{v}$:
$P = (3\hat{i} + 4\hat{j}) \cdot (6\hat{i} + 8\hat{j}) = (3 \times 6) + (4 \times 8) = 18 + 32 = 50$ W.
Question 2: Constant Power Kinematics
An automobile of mass $m$ accelerates from rest. If the engine supplies a constant power $P$, the velocity at time $t$ is given by:
(a) $\frac{Pt}{m}$
(b) $\sqrt{\frac{2Pt}{m}}$
(c) $\sqrt{\frac{Pt}{2m}}$
(d) $\frac{P^2 t}{m}$
Solution: (b)
From the derivation in Section 5:
$W_{net} = \Delta K$
Since power is constant, work done is $W = Pt$.
$Pt = \frac{1}{2}mv^2 – 0 \implies v = \sqrt{\frac{2Pt}{m}}$.
Question 3: Average Power of a Pump
A pump is required to lift 600 kg of water per minute from a well 25 m deep and eject it with a speed of 50 m/s. Calculate the power required to perform this task. (Take $g = 10$ m/s$^2$).
(a) 15.0 kW
(b) 12.5 kW
(c) 25.0 kW
(d) 30.5 kW
Solution: (a)
Mass lifted per second (mass flow rate) $dm/dt = \frac{600 \text{ kg}}{60 \text{ s}} = 10$ kg/s.
The pump does work against gravity and provides kinetic energy.
$P = \frac{dW}{dt} = \frac{d}{dt} \left(mgh + \frac{1}{2}mv^2\right)$
$P = \left(\frac{dm}{dt}\right)gh + \frac{1}{2}\left(\frac{dm}{dt}\right)v^2$
$P = (10)(10)(25) + \frac{1}{2}(10)(50)^2$
$P = 2500 + 5(2500) = 2500 + 12500 = 15000$ W = 15.0 kW.
Question 4: P-t Graph Interpretation
The power delivered to a particle of mass 2 kg varies with time as $P = 3t^2$ Watts. If the particle is at rest at $t = 0$, its velocity at $t = 2$ s will be:
(a) 2 m/s
(b) 4 m/s
(c) $2\sqrt{2}$ m/s
(d) $\sqrt{2}$ m/s
Solution: (c)
Work done is the integral of power: $W = \int_{0}^{2} P dt = \int_{0}^{2} 3t^2 dt = [t^3]_{0}^{2} = 8$ J.
By the Work-Energy Theorem, $W = \Delta K$:
$8 = \frac{1}{2} m v^2 – 0$
$8 = \frac{1}{2} (2) v^2 \implies v^2 = 8 \implies v = \sqrt{8} = 2\sqrt{2}$ m/s.
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