1.0 Concept of Work
In common language, any physical or mental exertion is called work. However, in Physics, Work has a very specific definition. Work is said to be done only when a force applied on a body makes the body move (i.e., there is a displacement in the direction of the force).
Scientific Definition of Work
Work done by a force is equal to the product of the magnitude of the force and the displacement of the body in the direction of the force.
- S.I. Unit: Joule ($J$). $1\,J = 1\,N \times 1\,m$.
- C.G.S. Unit: erg. ($1\,J = 10^7 \text{ erg}$).
- Nature: Scalar quantity.
Mathematical Expression for Work
$$W = F \cdot S \cos \theta$$
Where: $W$ = Work done, $F$ = Force, $S$ = Displacement, $\theta$ = Angle between Force and Displacement.
1.1 Special Cases of Work Done
The amount of work done depends heavily on the angle ($\theta$) at which the force is applied relative to the displacement:
- Positive Work ($\theta = 0^\circ$): Force and displacement are in the same direction. $\cos 0^\circ = 1$. Example: A coolie lifting a load.
- Zero Work ($\theta = 90^\circ$): Force is perpendicular to displacement. $\cos 90^\circ = 0$. Example: A coolie carrying a load on his head while walking on a horizontal road (Force of gravity is downward, displacement is horizontal).
- Negative Work ($\theta = 180^\circ$): Force acts opposite to the direction of displacement. $\cos 180^\circ = -1$. Example: Work done by the force of friction.
Work is zero in two conditions:
1. When displacement is zero ($S = 0$), even if force is large (e.g., pushing a wall).
2. When force is perpendicular to displacement (Centripetal force in circular motion is always perpendicular to velocity, so work done by centripetal force is always zero).
A force of $10\,N$ acts on a body and displaces it by $2\,m$ in a direction making an angle of $60^\circ$ with the direction of the force. Calculate the work done. ($\cos 60^\circ = 0.5$)
Solution:
1. Given: $F = 10\,N$, $S = 2\,m$, $\theta = 60^\circ$.
2. Formula: $W = F \cdot S \cos \theta$.
3. Calculation: $W = 10 \times 2 \times \cos 60^\circ$
4. $W = 20 \times 0.5 = \mathbf{10\,J}$.
Final Answer: The work done is $10\,J$.
James Prescott Joule, after whom the unit of work is named, was so obsessed with the concept of heat and work that he even spent his honeymoon measuring the temperature difference between the top and bottom of a waterfall!
2.0 Concept of Power
In Physics, doing work is not the only thing that matters; how fast the work is done is equally important. While "Work" depends only on the force and displacement, Power introduces the factor of time. If two people perform the same amount of work, the one who does it faster is said to have more power.
Definition of Power
Power is defined as the rate of doing work. It is a scalar quantity.
- S.I. Unit: Watt ($W$). $1\,W = 1\,J/s$.
- Other Units: kilowatt ($kW$), megawatt ($MW$), and Horsepower ($hp$).
- Conversion: $1\,hp = 746\,W$.
Mathematical Expression for Power
$$P = \frac{W}{t} = F \times v$$
Where: $P$ = Power, $W$ = Work done, $t$ = Time, $F$ = Force, $v$ = Average Velocity.
2.1 Distinction between Work and Power
It is crucial for ICSE students to distinguish between these two frequently confused terms:
| Feature | Work | Power |
|---|---|---|
| Definition | Product of Force and Displacement. | Rate of doing work. |
| Time Factor | Independent of time. | Depends on time. |
| S.I. Unit | Joule ($J$). | Watt ($W$). |
Always ensure time is in seconds when calculating Power in Watts. If the problem gives time in minutes or hours, convert it first!
Example: $2$ minutes = $120$ seconds.
A machine raises a load of $750\,N$ through a height of $16\,m$ in $5$ seconds. Calculate: (i) the work done by the machine and (ii) the power at which the machine works.
Solution:
1. Work Done: $W = F \times S = 750 \times 16 = \mathbf{12,000\,J}$.
2. Power: $P = \frac{W}{t} = \frac{12,000}{5} = \mathbf{2,400\,W}$.
Final Answer: Work = $12,000\,J$; Power = $2.4\,kW$.
The term "Horsepower" was created by James Watt to compare the output of steam engines with the power of draft horses. He determined that a horse could turn a mill wheel 144 times in an hour—leading to the calculation we still use for engines today!
3.0 Concept of Energy
Energy is a fundamental concept in Physics that represents the capacity to do work. A body that possesses energy can exert a force on another object to produce displacement. Energy and work are two sides of the same coin; the amount of energy a body possesses is equal to the amount of work it can perform.
Mechanical Energy
The energy possessed by a body due to its state of rest or its state of motion is called Mechanical Energy. It exists in two forms:
- Potential Energy ($U$): Energy due to position or configuration.
- Kinetic Energy ($K$): Energy due to the state of motion.
Unit: Like work, the S.I. unit of energy is the Joule ($J$).
3.1 Gravitational Potential Energy
When a body is raised above the ground, work is done against gravity. This work is stored in the body as Gravitational Potential Energy. The higher the body is lifted, the greater the potential energy it acquires.
Potential Energy Formula
$$U = mgh$$
Where: $m$ = mass, $g$ = acceleration due to gravity ($\approx 9.8\,m/s^2$), $h$ = vertical height.
3.2 Kinetic Energy
Every moving object possesses Kinetic Energy. To bring a moving object to rest, an equal amount of work must be done against its motion. Kinetic Energy is always positive because mass ($m$) is positive and velocity squared ($v^2$) is always positive.
Kinetic Energy Formula
$$K = \frac{1}{2}mv^2$$
Where: $m$ = mass of the body, $v$ = velocity of the body.
Potential energy is always relative. Usually, we take the ground level as zero potential energy ($h=0$). If a body is moved from height $h_1$ to $h_2$, the change in potential energy is $mg(h_2 - h_1)$. Always define your reference level clearly in numericals!
A ball of mass $0.5\,kg$ is slowed down from a velocity of $20\,m/s$ to $10\,m/s$. Calculate the change in its kinetic energy.
Solution:
1. Initial K.E. ($K_i$): $\frac{1}{2} \times 0.5 \times (20)^2 = 0.25 \times 400 = 100\,J$.
2. Final K.E. ($K_f$): $\frac{1}{2} \times 0.5 \times (10)^2 = 0.25 \times 100 = 25\,J$.
3. Change in K.E.: $K_f - K_i = 25 - 100 = \mathbf{-75\,J}$.
Final Answer: The kinetic energy decreased by $75\,J$. (The negative sign indicates loss of energy).
A compressed spring possesses Elastic Potential Energy. When you wind up a mechanical watch, you are doing work to compress a spring, storing energy that is then slowly released to move the clock hands!
4.0 Work-Energy Theorem
The Work-Energy Theorem provides a direct link between the work done on a body and the change in its kinetic energy. It is an essential tool for solving complex mechanics problems where finding acceleration or time might be difficult.
Statement of the Theorem
The work done by a force acting on a moving body is equal to the increase (or change) in its kinetic energy.
$$W = \Delta K = K_f - K_i$$
Relationship between K.E. and Momentum
Momentum ($p$) and Kinetic Energy ($K$) are related by:
$$K = \frac{p^2}{2m} \quad \text{or} \quad p = \sqrt{2mK}$$
This formula is frequently tested in ICSE MCQs regarding light vs. heavy bodies.
4.1 Forms of Energy and Conversions
Energy exists in various forms and can be transformed from one form to another. According to the Law of Conservation of Energy, the total energy of an isolated system remains constant.
| Device/Situation | Energy Transformation |
|---|---|
| Electric Motor | Electrical $\rightarrow$ Mechanical |
| Solar Cell | Light $\rightarrow$ Electrical |
| Battery (while in use) | Chemical $\rightarrow$ Electrical |
| Microphone | Sound $\rightarrow$ Electrical |
If two bodies of different masses have the same momentum, the lighter body will have more kinetic energy. Conversely, if they have the same kinetic energy, the heavier body will have more momentum. Use the formula $K = p^2/2m$ to verify this during exams!
Two bodies $A$ and $B$ have masses $m$ and $4m$ respectively. If they have the same kinetic energy, find the ratio of their linear momenta.
Solution:
1. Given: $K_A = K_B = K$. Masses: $m_A = m$, $m_B = 4m$.
2. From $p = \sqrt{2mK}$, we have:
$p_A = \sqrt{2mK}$ and $p_B = \sqrt{2(4m)K} = \sqrt{8mK} = 2\sqrt{2mK}$.
3. Ratio: $\frac{p_A}{p_B} = \frac{\sqrt{2mK}}{2\sqrt{2mK}} = \frac{1}{2}$.
Final Answer: The ratio of their momenta is 1:2.
The Law of Conservation of Energy means that the total amount of energy in the Universe has remained exactly the same since the Big Bang. We never "waste" energy; we simply convert it into non-useful forms like heat or sound (Energy Degradation).
5.0 Conservation of Mechanical Energy
The Law of Conservation of Mechanical Energy states that if there are no external forces like friction or air resistance acting on a system, the sum of Potential Energy ($U$) and Kinetic Energy ($K$) remains constant. Energy may transform from one form to another, but the total mechanical energy is conserved.
Mechanical Energy Conservation
For a body under the influence of conservative forces (like gravity):
$$E_{total} = K + U = \text{Constant}$$
5.1 Verification: Freely Falling Body
Consider a body of mass $m$ falling from a height $h$. Let's analyze the energy at three points: A (Top), B (Middle), and C (Ground).
- At Point A (Height $h$): Velocity is $0$, so $K = 0$. Potential Energy $U = mgh$. Total Energy = $\mathbf{mgh}$.
- At Point B (Distance $x$ from top): Potential Energy $U = mg(h-x)$. Using $v^2 = 2gx$, Kinetic Energy $K = \frac{1}{2}m(2gx) = mgx$. Total Energy = $mg(h-x) + mgx = \mathbf{mgh}$.
- At Point C (Ground): Height is $0$, so $U = 0$. Velocity $v^2 = 2gh$, so $K = \frac{1}{2}m(2gh) = mgh$. Total Energy = $\mathbf{mgh}$.
Velocity of a Falling Body
The velocity ($v$) of a body falling from height ($h$) just before hitting the ground is:
$$v = \sqrt{2gh}$$
In the real world, mechanical energy is not perfectly conserved. A falling object loses some energy as heat due to air friction. The "Law of Conservation of Energy" still holds for total energy, but mechanical energy decreases.
A ball of mass $200\,g$ is dropped from a height of $5\,m$. Find its kinetic energy just before it hits the ground. ($g = 10\,m/s^2$)
Solution:
1. Initial Potential Energy (at top): $U = mgh = 0.2\,kg \times 10 \times 5 = \mathbf{10\,J}$.
2. According to Conservation of Energy: Total P.E. at the top converts entirely into K.E. at the bottom (assuming no friction).
3. Therefore, Final Kinetic Energy: $\mathbf{10\,J}$.
Final Answer: Kinetic energy at the ground is $10\,J$.
Hydroelectric power plants use this exact principle! They store water at a high elevation (Maximum P.E.). As the water falls, it gains K.E., which is then used to spin turbines and generate electrical energy.