The idea of work as a form of energy transfer dates back to ancient times, but it wasn’t until the 17th and 18th centuries that the concept began to take a more scientific form.

Aspects of the concept of work done appear as early as 60 AD in the writings of Hero of **Alexandria.** He observed that if a weight were raised using a pulley system by exerting a force less than the weight being lifted, the rope must be pulled faster than the weight rises.

**Galileo** hinted at energy conservation without explicitly discussing it, suggesting that certain machines were impossible because they would cheat nature. In his 1664 “Principia Philosophiae,” Descartes proposed laws of collisions and introduced the concept of quantity of motion, calculated as the product of mass and undirected speed.

**Leibniz **disagreed with Descartes’ notion of conservation of motion. He distinguished between “vis viva” (living force), defined as mass times velocity squared (akin to modern kinetic energy), and “vis mortua” (dead force), related to potential energy. Leibniz argued that “vis viva” was conserved, a concept debated for over 50 years among philosophers.

The term “work” was introduced in 1826 by the French mathematician **Gaspard-Gustave Coriolis. He described work as a **“weight lifted through a height,” which related to the use of steam engines to lift water out of mines. Coriolis gave the correct formula for change in kinetic energy associated with work.

This historical perspective shows that the concept of work evolved from practical observations to a fundamental scientific principle that conserves energy. It’s a journey from the simple mechanics of pulleys to the sophisticated laws of motion and energy conservation.

Table of Contents

**What is Work?**

In physics, “work” is a measure of the energy transferred to or from an object using a force acting on the object as it moves through a distance. It’s important to understand that work is only done when a force causes displacement or movement in the direction of the force applied.

Think of it this way: when you push or pull an object, you’re exerting a force on it. If that object moves in response to your force, then work is being done. However, if you push against a wall and the wall doesn’t move, then no work is being done because there’s no displacement.

So, work involves both force and displacement. The greater the force applied and the greater the distance over which the force is applied, the more work is done.

This concept is crucial in understanding various phenomena in physics, including the transfer and transformation of energy in systems.

**Work Formula**

To calculate the work done, we use a simple formula:

\(\displaystyle W = F \cdot d \cdot \cos(\theta) \)

**W**is the work done (measured in Joules).**F**is the force applied (measured in Newtons).**d**is the distance the object moves (measured in meters).**(cos(θ)**is the cosine of the angle between the force applied and the direction of movement.

**Unit of Work**

When we talk about the “Unit of Work” in physics, we’re referring to the standard measurement used to express the amount of work done. It’s like how we use meters for distance and seconds for time.

**S.I. Unit:**

- The
**S.I. Unit**for work is the**Joule (J)**. It’s named after the English physicist James Prescott Joule. **One Joule**is defined as the amount of work done when a force of**one Newton**is applied to an object, moving it**one meter**in the direction of the force.- So, if you push a book with a force of one Newton for a distance of one meter, you’ve done one Joule of work.

**C.G.S Unit:**

- In the
**C.G.S system**(which stands for centimeters, grams, seconds), the unit of work is the**erg**. **One erg**is much smaller than a Joule. It’s the work done by a force of**one dyne**(which is one gram-force) moving an object**one centimeter**.- To put it in perspective, one Joule is equal to
**10^7 ergs**. That’s ten million ergs!

We have different units because scientists around the world used different systems historically. The S.I. system is now the most widely used because it’s based on the metric system, which is easier to understand and more consistent.

The Joule can also be related to other forms of energy, like heat. For example, one Joule is also the amount of heat required to raise the temperature of one gram of water by 0.24 degrees Celsius.

**Dimensional Formula of Work:**

The dimensional formula is a way to express the units of a physical quantity in terms of the basic dimensions of mass (M), length (L), and time (T). For work, this formula is crucial because it helps us understand how work relates to these fundamental units.

\(\displaystyle [M^1L^2T^{-2}] \)

To derive the dimensional formula of work, we start with the basic definition of work. Work (W) is the product of force (F) and displacement (d). So, we can write:

\(\displaystyle W = F \times d \)

Now, we know that force itself is derived from mass (m) and acceleration (a). The formula for force is:

\(\displaystyle F = m \times a \)

Acceleration is the rate of change of velocity, which is speed over time. So, its formula is:

\(\displaystyle a = \frac{v}{t} \)

But velocity (v) is just displacement over time. So, we replace ‘v’ with ‘d/t’, which gives us:

\(\displaystyle a = \frac{d}{t^2} \)

Now, we substitute ‘a’ back into the formula for force:

\(\displaystyle F = m \times \frac{d}{t^2} \)

This gives us the dimensions of force as:

\(\displaystyle [F] = [M] \times \left[\frac{L}{T^2}\right] \)

So, the dimensional formula for force is:

\(\displaystyle [F] = [M^1L^1T^{-2}] \)

Now, we go back to our original work formula. Since work is force times displacement, we multiply the dimensions of force by the dimensions of displacement (L):

\(\displaystyle [W] = [F] \times [L] \)

Substituting the dimensions of force, we get:

\(\displaystyle [W] = [M^1L^1T^{-2}] \times [L^1] \)

This simplifies to:

\(\displaystyle [W] = [M^1L^2T^{-2}] \)

The dimensional formula of work is:

\(\displaystyle [M^1L^2T^{-2}] \)

This tells us that work is related to mass, length squared, and the inverse of time squared.

**Types of Work**

In physics, when we talk about the types of work, we’re referring to how the direction of the force applied to an object relates to the direction the object moves. Depending on this relationship, work can be classified into three main types:

**Positive Work**

**Positive Work**happens when the force applied to an object and the displacement of the object is in the same direction.- For example, if you push a car and it moves forward, the work you’re doing on the car is positive because the car is moving in the direction you’re pushing.

**Negative Work**

**Negative Work**occurs when the force applied to an object is in the opposite direction of its displacement.- An example of negative work is when you’re pulling back on a dog’s leash, but the dog is trying to run forward. The work you’re doing on the dog is negative because the force you’re applying (pulling back) is opposite to the dog’s movement (running forward).

**Zero Work**

**Zero Work**is a bit different. It happens when the force applied to an object is perpendicular to the direction of displacement, or when there is no displacement at all.- Imagine holding a heavy bag still. Even though you’re applying force to keep the bag from falling, you’re not moving it. So, the work done is zero because there’s no displacement.

These types of work help us understand how forces interact with objects to cause movement or prevent it. They’re fundamental concepts that show up everywhere in physics, from simple machines to complex systems. Remember, the key to work is displacement—if there’s no movement, there’s no work, regardless of how much force is applied!

**Factors Affecting Work**

When we talk about the work done by a force in physics, it’s not just about how hard you push or pull something. There are specific factors that determine whether work is done and how much work is done. These factors are:

**Force:**

**Force**is the push or pull that can cause an object with mass to change its velocity (speed up, slow down, or change direction).- It’s a vector quantity, which means it has both magnitude (how strong it is) and direction (which way it’s going).
- If the force acting on an object is zero, no matter what the object is doing, the work done by that force is also zero.

**Displacement**:

**Displacement**is the distance moved in a specific direction. It’s also a vector quantity.- For work to be done, there must be displacement in the direction of the force. If you push against a wall with all your might and it doesn’t move, then no work has been done on the wall.

**Angle Between Force and Displacement:**

- The angle between the force applied and the direction of displacement is crucial.
- Work can be
**positive**,**negative**, or**zero**depending on this angle:- If the force and displacement are in the same direction, the work is positive.
- If they’re in opposite directions, the work is negative.
- If the force is perpendicular to the displacement, no work is done (zero work).

These factors are essential in determining the work done in any situation.

**Work done by a constant force**

When a constant force acts on an object, the work done is the product of the force and the displacement in the direction of the force.

When we say a constant force is doing work on an object, it means that the force is not changing in magnitude or direction as the object moves. The work done by this constant force can be calculated easily.

Let’s consider a force \(\displaystyle\vec{F} \) acting on an object, causing it to move a displacement \(\displaystyle\vec{d}\) in a straight line. The angle between the force and the displacement is (θ).

The work done (W) by the constant force is given by the formula:

\(\displaystyle W = F \cdot d \cdot \cos(\theta) \)

Here, (F) is the magnitude of the force, (d) is the magnitude of the displacement, and (\(\displaystyle\cos(\theta)\) ) is the cosine of the angle between the force and the displacement vector.

Now, if the force is constant and acts along the direction of displacement, then (θ= 0 ) and (cos(0) = 1 ). So the formula simplifies to:

\(\displaystyle W = F \cdot d \)

This means that the work done by a constant force is just the product of the magnitude of the force and the magnitude of the displacement in the direction of the force.

**Example**: Imagine you’re pushing a shopping cart with a constant force across a flat parking lot. If you push the cart with a force of 10 Newtons for a distance of 5 meters, the work done by you on the cart is:

\(\displaystyle W = 10 \, \text{N} \times 5 \, \text{m} = 50 \, \text{J} \)

So, you’ve done 50 Joules of work on the shopping cart.

This concept is fundamental in understanding how energy is transferred by forces in physics. It’s a simple yet powerful idea that shows how even constant forces can do varying amounts of work depending on the distance over which they act.

**Work done by a variable force**

For a variable force, the work done is calculated using integration, considering the force at each point along the object’s path.

When a force changes its magnitude or direction as it acts on an object, we call it a variable force. Unlike a constant force, calculating work done by a variable force requires considering the force at each point of the object’s path.

To understand how to calculate work done by a variable force,

let’s consider a force (F) that varies with displacement (x). The work done (W) by this force over a small displacement (∆x) is approximately the force at that point times the small displacement, which is (∆W = F(x) .∆*x*)

However, as the displacement (∆x) becomes very small, we need to add up (integrate) all these small amounts of work done over the entire path of the displacement. Mathematically, this is expressed as:

\(\displaystyle W = \int_{x_i}^{x_f} F(f) \, dx \)

where (*x*_{i}) is the initial position and (x_{f}) is the final position.

**Example with a Spring**: Consider a spring that follows Hooke’s Law, where the force (F) is proportional to the displacement (x) from its equilibrium position and is given by (F = -kx), where (k) is the spring constant.

The negative sign indicates that the force exerted by the spring is opposite to the direction of displacement. To find the work done by this variable force as the spring is stretched or compressed, we integrate the force over the displacement:

\(\displaystyle W = \int_{x_i}^{x_f} (-kx) \, dx \)

This integral evaluates to:

\(\displaystyle W = -\frac{1}{2}kx^2 \Big|_{x_i}^{x_f} \)

which gives us the work done by the spring force over the displacement from (*x*_{i}) to (*x*_{f}).

This concept of work done by a variable force is crucial in understanding real-world scenarios where forces are not constant, such as in the stretching of springs, the motion of planets under varying gravitational forces, or the deceleration of vehicles.

**Derivation of the Work Done Formula**

The work-energy theorem states that the work done by all forces acting on a particle equals the change in its kinetic energy. The derivation involves equating the net work done to the change in kinetic energy, which can be represented as:

\(\displaystyle W = \Delta KE \)

In physics, the work done by a force is related to the energy transferred to an object. To derive the formula for work done, we’ll start with the basic definition and then connect it to the change in kinetic energy of the object.

Work done (W) is defined as the product of the force (F) applied to an object and the displacement (d) of the object in the direction of the force. Mathematically, it’s given by:

\(\displaystyle W = F \cdot d \cdot \cos(\theta) \)

where (θ) is the angle between the force and the displacement.

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy (KE). If an object starts with an initial velocity (u) and ends with a final velocity (v), the change in kinetic energy is:

\(\displaystyle \Delta KE = \frac{1}{2}m(v^2 – u^2) \)

Now, let’s derive the work-done formula using the work-energy theorem. We know that the final velocity (v) can be related to the initial velocity (u), acceleration (a), and displacement (d) by the kinematic equation:

\(\displaystyle v^2 = u^2 + 2ad \)

Substituting this into the change in kinetic energy, we get:

\(\displaystyle \Delta KE = \frac{1}{2}m(2ad) \)

Simplifying, we find:

\(\displaystyle \Delta KE = mad \)

Since the work done is equal to the change in kinetic energy, we have:

\(\displaystyle W = mad \)

But, according to Newton’s second law, force (F) is the product of mass (m) and acceleration (a):

\(\displaystyle F = ma \)

So, substituting ( F ) into the work done equation, we get:

\(\displaystyle W = Fd \)

This is the work done by a force when it acts in the direction of the displacement.

This derivation shows that work is not just about applying a force; it’s about how that force moves an object and changes its energy. It’s a fundamental concept that links forces, motion, and energy in the physical world.

By understanding this derivation, students can better grasp how energy is conserved and transferred in various physical processes.

**Also Read: Position, Path Length And Displacement**

**Solved Example**

**Example 1: A force of 20 N is applied to push a block horizontally for a distance of 5 meters. If the force is applied at an angle of 30 degrees to the horizontal, calculate the work done.**

**Solution: **Work done by a force at an angle is given by the formula:

\(\displaystyle W = F \cdot d \cdot \cos(\theta) \)

Given (F = 20) N, (d = 5) m, and (θ = 30^{∘}):

\(\displaystyle W = 20 \times 5 \times \cos(30^\circ) \)

\(\displaystyle W = 20 \times 5 \times \frac{\sqrt{3}}{2} \)

\(\displaystyle W = 50\sqrt{3} \, \text{J} \)

**Example 2: A force of 30 N is used to lift a box vertically upwards for a distance of 4 meters. Calculate the work done against gravity.**

**Solution:** Work done against gravity is given by the formula:

\(\displaystyle W = F \cdot d \cdot \cos(\theta) \)

In this case, the angle (θ) between the force and the displacement is ( 0^{∘}), as the force and displacement are in the same direction.Given (F = 30) N, (d = 4) m, and (θ= 0^{∘}):

\(\displaystyle W = 30 \times 4 \times \cos(0^\circ) \)

\(\displaystyle W = 30 \times 4 \times 1 \)

\(\displaystyle W = 120 \, \text{J} \)

**Example 3: A spring with a spring constant of 200 N/m is compressed by 0.1 meters. Calculate the work done to compress the spring.**

**Solution:** Work done to compress a spring is given by the formula:

\(\displaystyle W = \frac{1}{2} k x^2 \)

Given (k = 200) N/m and (x = 0.1) m:

\(\displaystyle W = \frac{1}{2} \times 200 \times (0.1)^2 \)

\(\displaystyle W = 1 \, \text{J} \)

**Example 4: A person pulls a cart horizontally with a force of 50 N. The cart moves a distance of 8 meters. If the force is in the opposite direction to the displacement, calculate the work done.**

**Solution:** When the force is opposite to the direction of displacement, the work done is negative. The formula for work done remains the same:

\(\displaystyle W = F \cdot d \cdot \cos(\theta) \)

However, in this case, (θ= 180^{∘}). Given (F = 50) N, (d = 8) m,

\(\displaystyle W = 50 \times 8 \times \cos(180^\circ) \)

\(\displaystyle W = 50 \times 8 \times (-1) \)

\(\displaystyle W = -400 \, \text{J}\)

**Example 5: A particle moves along the x-axis under the influence of a force (F = -3x ^{2}) N. If it moves from (x_{i} = 1) m to (x_{f} = 3) m, calculate the work done by the force.**

**Solution:** Given (F(x) = -3x^{2}), and the limits of integration are (x_{i} =1) m and (x* _{f}* =3 ) m.

\(\displaystyle W = \int_{1}^{3} -3x^2 \, dx \)

\(\displaystyle W = \left[ -x^3 \right]_{1}^{3} \)

\(\displaystyle W = -(3^3 – 1^3) \, \text{J} = -26 \, \text{J} \)

**Example 6: A particle of mass 2 kg is pulled along a straight line by a force (F = 3x) N, where (x) is in meters. If the particle is displaced from (x _{i} = 1) m to (x_{f} = 4) m, calculate the work done by the force.**

**Solution: **Given (F(x) = 3x), and the limits of integration are (x* _{i}* = 1) m and (x

_{f}= 4) m.

\(\displaystyle W = \int_{1}^{4} 3x \, dx \)

\(\displaystyle W = \left[ \frac{3}{2}x^2 \right]_{1}^{4} \)

\(\displaystyle W = \frac{3}{2}(4^2 – 1^2) \, \text{J} = 21 \, \text{J} \)

**FAQs**

**What is work in physics, and how is it different from everyday usage of the term?**

In physics, work is defined as the product of force and displacement in the direction of the force. It represents the transfer of energy from one object to another by the application of force over a distance. This definition differs from the everyday usage of the term, where work often implies effort or activity.

**Can you explain the significance of the dot product in the formula for work?**

The dot product in the formula for work, \(\displaystyle W = \vec{F} \cdot \vec{d} \), ensures that only the component of force acting in the direction of displacement contributes to the work done. This is because work involves the transfer of energy in the direction of motion, so only the force component parallel to the displacement is considered.

**Is work done only when an object moves a certain distance?**

Yes, work is done only when there is a displacement of an object in the direction of the applied force. If the force is applied but the object doesn’t move, no work is done. Similarly, if the force and displacement are perpendicular to each other, the work done is zero.

**How does the angle between force and displacement affect the work done?**

The angle between the force and displacement vectors determines the amount of work done. When the force and displacement are parallel (angle = 0°), the work done is maximum. As the angle increases to 90° (perpendicular), the work done decreases until it becomes zero. If the angle is greater than 90°, the work done becomes negative.

**Can you provide examples of different types of work?**

Examples of work include lifting a box against gravity (work done against gravity), pushing a car along a flat road (work done to overcome friction), and stretching a spring (work done to store elastic potential energy). In each case, force is applied over a distance, resulting in work done.

**How does work relate to kinetic energy?**

Work is directly related to kinetic energy through the work-energy theorem, which states that the work done on an object equals the change in its kinetic energy. When work is done on an object, it gains or loses kinetic energy depending on whether the work is done by an external force or against it.

**Can you explain the concept of negative work?**

Negative work occurs when the force applied to an object opposes its motion or displacement. This results in a decrease in the object’s kinetic energy or potential energy. For example, when friction slows down a moving object, the work done by friction is negative because it acts in the direction opposite to the displacement.