Work and Energy - Scientific concept of work done, kinetic energy, potential energy, and the law of conservation of energy

Have you ever worked hard studying for an exam for four hours straight, feeling completely exhausted, only for a physicist to tell you, "Technically, you did zero work!"?

Sounds unfair, right? But in the world of physics, terms like Work and Energy have very specific, precise scientific definitions. Today, we are going to explore these exciting concepts step-by-step as outlined in your NCERT Class 9 Science syllabus!

1. Scientific Concept of Work Done

In everyday life, mental strain, standing at a bus stop holding a heavy bag, or pushing against an unyielding wall are all considered "hard work." However, in science, these do not count as work!

The Two Golden Conditions for Work

For Work to be scientifically done, two mandatory conditions must be satisfied:

1. A force must act on an object.
2. The object must get displaced (move) in the direction (or opposite direction) of the force.

$\text{Work Done } (W) = \text{Force } (F) \times \text{Displacement } (s)$

$W = F \times s$

Units of Work

• SI Unit of Force: Newton ($\text{N}$)
• SI Unit of Displacement: Metre ($\text{m}$)
• SI Unit of Work: Newton-metre ($\text{N m}$) or Joule ($\text{J}$)

1 Joule Definition: One Joule of work is done on an object when a force of $1\text{ N}$ displaces it by $1\text{ m}$ along the line of action of the force.

Types of Work Done

Work can be Positive, Negative, or Zero:

2. Kinetic Energy (Energy in Motion)

What gives a fast-moving cricket ball the power to break a window glass? It’s Kinetic Energy!

What is Kinetic Energy?

Kinetic Energy ($E_k$) is the energy possessed by an object due to its motion. Any object that is moving—a flying airplane, running water, a rolling bowling ball—has kinetic energy.

Mathematical Derivation of Kinetic Energy

Let an object of mass $m$ move with initial velocity $u$. Let a constant force $F$ displace it by distance $s$, changing its velocity to $v$ with an acceleration $a$.

From the 3rd equation of motion: $v^2 - u^2 = 2as \implies s = \frac{v^2 - u^2}{2a}$

From Newton’s Second Law of Motion: $F = m \times a$

Since Work Done $W = F \times s$: $W = (m \times a) \times \left(\frac{v^2 - u^2}{2a}\right) = \frac{1}{2} m (v^2 - u^2)$

If the object starts from rest ($u = 0$): $W = \frac{1}{2} m v^2$

Since the work done on the object equals its kinetic energy gained: $E_k = \frac{1}{2} m v^2$

Key Takeaway: Kinetic energy is directly proportional to mass ($m$) and the square of velocity ($v^2$). If you double the speed of a vehicle, its kinetic energy increases by four times!

3. Potential Energy (Energy of Position or Shape)

Imagine pulling the string of a bow. As long as you hold the stretched string, nothing moves, yet it holds immense power. The moment you release it, the arrow flies away at high speed! Where did that energy come from? It was stored as Potential Energy.

What is Potential Energy?

Potential Energy ($E_p$) is the energy possessed by an object due to its position or change in shape/configuration.

• Elastic Potential Energy: Stored due to deformation (e.g., stretched rubber band, compressed spring).
• Gravitational Potential Energy: Stored due to an object's height above the ground.

Formula for Gravitational Potential Energy

When an object of mass $m$ is raised to a height $h$ against gravity ($g$):

• Minimum force required = Weight of object = $m \times g$
• Displacement = $h$

$\text{Work Done } (W) = \text{Force} \times \text{Displacement} = m \cdot g \cdot h$

$E_p = m g h$

(where $g \approx 9.8 \text{ m/s}^2$ or $10 \text{ m/s}^2$)

4. Law of Conservation of Energy

This is one of the most fundamental laws in all of physics!

Statement

Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system always remains constant.

$\text{Total Mechanical Energy} = \text{Kinetic Energy } (E_k) + \text{Potential Energy } (E_p) = \text{Constant}$

Verification using a Free-Falling Object

Consider a ball of mass $m$ dropped from a height $h$ above the ground:

1. At the Topmost Point (Height $h$):

   • Velocity $v = 0 \implies E_k = 0$
   • Potential Energy $E_p = mgh$
   • Total Energy $= 0 + mgh = \mathbf{mgh}$

2. At a Middle Point (falling through distance $x$, height remaining $h-x$):

   • Using $v^2 = u^2 + 2gx \implies v^2 = 2gx$
   • $E_k = \frac{1}{2} m (2gx) = mgx$
   • $E_p = mg(h - x) = mgh - mgx$
   • Total Energy $= mgx + mgh - mgx = \mathbf{mgh}$

3. Just Above the Ground (Height $0$, distance fallen $h$):

   • $v^2 = 2gh$
   • $E_k = \frac{1}{2} m (2gh) = mgh$
   • $E_p = 0$
   • Total Energy $= mgh + 0 = \mathbf{mgh}$

As the ball falls, its potential energy continuously converts into kinetic energy, but the sum of both remains constant at every point!

Practice Questions with Detailed Solutions

Let's test your understanding with three classic NCERT numerical problems!

Question 1 (Work Done)

A force of $7\text{ N}$ acts on an object. The displacement is $8\text{ m}$ in the direction of the force. What is the work done in this case?

Solution:

• Given:
  • Force ($F$) = $7\text{ N}$
  • Displacement ($s$) = $8\text{ m}$
• Formula: $W = F \times s$
• Calculation: $W = 7\text{ N} \times 8\text{ m} = 56\text{ J}$
• Answer: The work done on the object is $56\text{ Joules}$.

Question 2 (Kinetic Energy)

An object of mass $15\text{ kg}$ is moving with a uniform velocity of $4\text{ m/s}$. What is the kinetic energy possessed by the object?

Solution:

• Given:
  • Mass ($m$) = $15\text{ kg}$
  • Velocity ($v$) = $4\text{ m/s}$
• Formula: $E_k = \frac{1}{2} m v^2$
• Calculation: $E_k = \frac{1}{2} \times 15 \times (4)^2$ $E_k = \frac{1}{2} \times 15 \times 16$ $E_k = 15 \times 8 = 120\text{ J}$
• Answer: The kinetic energy possessed by the object is $120\text{ Joules}$.

Question 3 (Potential Energy & Energy Conservation)

Find the potential energy of an object of mass $10\text{ kg}$ raised to a height of $6\text{ m}$ above the ground. (Take $g = 9.8\text{ m/s}^2$). Also, state its kinetic energy just before hitting the ground if it is allowed to fall freely.

Solution:

• Part 1: Potential Energy at Height $h$

  • Given: $m = 10\text{ kg}$, $h = 6\text{ m}$, $g = 9.8\text{ m/s}^2$
  • Formula: $E_p = m \times g \times h$
  • Calculation: $E_p = 10 \times 9.8 \times 6 = 588\text{ J}$

• Part 2: Kinetic Energy just before impact

  • According to the Law of Conservation of Energy, all potential energy at the maximum height converts into kinetic energy just before touching the ground.
  • Therefore, Kinetic Energy ($E_k$) near ground = Initial Potential Energy ($E_p$) at top = $588\text{ Joules}$.

• Answer:

  • Potential Energy at top = $588\text{ J}$
  • Kinetic Energy just before hitting ground = $588\text{ J}$

Summary Checklist for Revision

1. Work ($W = F \cdot s$) requires both force and displacement. SI unit: Joule ($\text{J}$).
2. Kinetic Energy ($E_k = \frac{1}{2}mv^2$) depends on mass and velocity square.
3. Potential Energy ($E_p = mgh$) depends on mass, gravity, and height.
4. Law of Conservation of Energy: Total Energy $= E_k + E_p =$ Constant.

Keep practicing your formulas, solve NCERT exemplar problems, and keep observing the physics around you every day. Happy learning!