Work Energy and Power for Class 11 with Concepts and Examples

Work is one of the most basic and important concepts in Physics. In simple terms, work is said to be done when a force acts on a body and the body undergoes a displacement. If force is applied but the object does not move, then no work is done in the Physics sense.

This concept helps students understand how force transfers energy to an object. It also forms the base for topics like kinetic energy, potential energy, power, and conservation of mechanical energy.

The mathematical expression for work is:

$W = \vec{F} \cdot \vec{s}$

or

$W = Fs\cos\theta$

where:
$W$ = work done
$F$ = magnitude of force
$s$ = displacement
$\theta$ = angle between force and displacement

Important Points About Work

• Work is a scalar quantity.

• The SI unit of work is joule $(J)$.

• Work depends on both force and displacement.

• Only the component of force in the direction of displacement does work.

Types of Work

There are majorly 3 types of work:

1. Positive Work: Work is positive when force and displacement are in the same direction.

Example: Gravity does positive work on a falling object.

2. Negative Work: Work is negative when force acts opposite to the direction of displacement.

Example: Friction does negative work on a moving body.

3. Zero Work: Work is zero when there is no displacement or when force is perpendicular to displacement

Example: In circular motion, centripetal force does zero work because it acts perpendicular to motion.

Work Done by a Constant Force

When a force remains unchanged in magnitude as well as direction during the motion of a body, it is called a constant force. In such cases, the calculation of work becomes simple because the same force acts throughout the displacement.

If a body moves through a displacement $\vec{s}$ under the action of a constant force $\vec{F}$, then the work done is given by:

$W = \vec{F} \cdot \vec{s}$

or

$W = Fs\cos\theta$

where:
$F$ = magnitude of force
$s$ = displacement
$\theta$ = angle between force and displacement

This formula shows that only the component of force along the direction of motion contributes to work done.

Understanding the Formula

The term $\cos\theta$ is very important because force and displacement may not always be in the same direction.

• If force acts along displacement, maximum work is done.

• If force acts opposite to displacement, work becomes negative.

• If force acts perpendicular to displacement, no work is done.

So, work done depends not only on the force and displacement but also on the angle between them.

Special Cases of Work Done by a Constant Force

1. When $\theta = 0^\circ$

If force and displacement are in the same direction:

$W = Fs\cos 0^\circ$

$W = Fs$

This is the maximum positive work possible for a given force and displacement.

2. When $\theta = 90^\circ$

If force is perpendicular to displacement:

$W = Fs\cos 90^\circ$

$W = 0$

So, no work is done.

3. When $\theta = 180^\circ$

If force acts opposite to displacement:

$W = Fs\cos 180^\circ$

$W = -Fs$

So, the work done is negative.

Physical Meaning of Positive and Negative Work

Positive work means energy is being transferred to the body, so the body may speed up.

Negative work means energy is being taken away from the body, so the body may slow down.

This idea becomes very important later in kinetic energy and the work-energy theorem.

Examples from Daily Life

• A person lifting a bag upward does positive work on the bag.

• Brakes applied to a moving bicycle do negative work.

• A person carrying a load on a horizontal road does zero work on the load in the vertical direction if the applied force is vertical and displacement is horizontal.

Example 1

A force of $15,N$ acts on a body and moves it by $4,m$ in the same direction. Find the work done.

Solution

Since force and displacement are in the same direction, $\theta = 0^\circ$

$W = Fs\cos\theta$

$W = 15 \times 4 \times \cos 0^\circ$

$W = 60,J$

Answer

Work done = $60,J$

Example 2

A force of $20,N$ acts on a body moving $5,m$, and the angle between force and displacement is $60^\circ$. Find the work done.

Solution

$W = Fs\cos\theta$

$W = 20 \times 5 \times \cos 60^\circ$

$W = 100 \times \frac{1}{2}$

$W = 50,J$

Answer

Work done = $50,J$

Example 3

A retarding force of $8,N$ acts on a moving object and brings it to rest over a distance of $3,m$. Find the work done by the retarding force.

Solution

Since the retarding force acts opposite to displacement, $\theta = 180^\circ$

$W = Fs\cos 180^\circ$

$W = 8 \times 3 \times (-1)$

$W = -24,J$

Answer

Work done = $-24,J$

Work Done by a Variable Force

In many real-life situations, force does not remain constant throughout the motion of a body. Its magnitude may change, its direction may change, or both may change continuously with position or time. In such cases, the simple formula used for constant force cannot be applied directly to the entire motion.

This type of force is called a variable force, and the method of finding work done by it is slightly different.

What is a Variable Force?

A variable force is a force whose magnitude or direction changes during motion.

Examples of variable force include:

• spring force

• gravitational force over large distances

• resistive force in some cases

• force acting on a body along a curved path

Since the force is not the same at every point, we divide the motion into very small displacements and calculate the small work done over each part.

Small Work Done Over a Small Displacement

Suppose a body moves by a very small displacement $d\vec{r}$ under the action of a force $\vec{F}$.

Then the small work done is:

$dW = \vec{F} \cdot d\vec{r}$

or

$dW = F,dr\cos\theta$

where $\theta$ is the angle between force and the small displacement.

This expression gives the infinitesimal work done.

Total Work Done by a Variable Force

To find the total work done from one position to another, all these very small work contributions are added using integration:

$W = \int \vec{F} \cdot d\vec{r}$

This is the general formula for work done by a variable force.

It means that work done is found by integrating the component of force along the direction of displacement over the entire path.

Variable Force in One Dimension

If the body moves only along the x-axis and force also acts along the x-direction, then the formula becomes simpler:

$W = \int_{x_1}^{x_2} F(x),dx$

Here, $F(x)$ means force is changing with position.

This form is commonly used in numerical problems.

Why Integration is Needed

For constant force, we can take force out as a fixed value and use:

$W = Fs\cos\theta$

But for variable force, force changes from point to point. So we cannot multiply one single value of force by the whole displacement. Instead, we take tiny elements of displacement and add their contributions.

That is why integration is needed.

Graphical Interpretation of Work Done by a Variable Force

If a graph is drawn between force and displacement, then the work done by a variable force is equal to the area under the force-displacement graph between the given limits.

This is one of the most important interpretations in this chapter.

Important Graph Points

• Area above the x-axis gives positive work

• Area below the x-axis gives negative work

• Net work done is the algebraic sum of all areas

So in graphical questions, students must be careful with signs.

Work Done by a Spring Force

A spring provides one of the most important examples of variable force.

According to Hooke’s law:

$F = -kx$

where:
$k$ = spring constant
$x$ = extension or compression

The negative sign shows that spring force always acts opposite to displacement.

If the spring is stretched or compressed from $x_1$ to $x_2$, then the work done by the spring is:

$W = \int_{x_1}^{x_2} (-kx),dx$

$W = -k \int_{x_1}^{x_2} x,dx$

$W = -k \left[\frac{x^2}{2}\right]_{x_1}^{x_2}$

$W = \frac{1}{2}k(x_1^2 - x_2^2)$

This is the work done by the spring.

If the question asks for work done on the spring, then the sign becomes opposite:

$W = \frac{1}{2}k(x_2^2 - x_1^2)$

This difference is very important in exam questions.

Example 1

A force acting on a particle varies as:

$F(x) = 2x$

Find the work done in moving the particle from $x = 1,m$ to $x = 3,m$.

Solution

Using:

$W = \int_{1}^{3} 2x,dx$

$W = 2\int_{1}^{3} x,dx$

$W = 2\left[\frac{x^2}{2}\right]_{1}^{3}$

$W = \left[x^2\right]_{1}^{3}$

$W = 9 - 1$

$W = 8,J$

Answer

Work done = $8,J$

Example 2

A force varies with displacement as:

$F = 3x + 1$

Find the work done from $x = 0$ to $x = 2,m$.

Solution

$W = \int_{0}^{2} (3x + 1),dx$

$W = \left[\frac{3x^2}{2} + x\right]_{0}^{2}$

$W = \left[\frac{3(4)}{2} + 2\right]$

$W = 6 + 2$

$W = 8,J$

Answer

Work done = $8,J$

Example 3

A spring of spring constant $k$ is stretched from natural length to a distance $x$. Find the work done on the spring.

Solution

For work done on the spring:

$W = \frac{1}{2}kx^2$

Answer

Work done on the spring = $\frac{1}{2}kx^2$

Key Differences Between Constant and Variable Force

Constant Force

• Magnitude and direction remain unchanged

• Work is found using $W = Fs\cos\theta$

Variable Force

• Magnitude or direction changes during motion

• Work is found using integration

This distinction is very important for clear understanding.

Kinetic Energy

Kinetic energy is the energy possessed by a body due to its motion. Whenever an object is moving, it has the ability to do work because of its speed. A stationary object has no kinetic energy, but the moment it starts moving, it gains kinetic energy.

This is one of the most important concepts in the chapter because it directly connects motion with energy. It also forms the base for understanding the work-energy theorem, collisions, and conservation of mechanical energy.

Definition of Kinetic Energy

Kinetic energy is defined as the energy possessed by a body by virtue of its motion.

If a body of mass $m$ moves with speed $v$, then its kinetic energy is:

$K = \frac{1}{2}mv^2$

where:
$m$ = mass of the body
$v$ = speed of the body

Meaning of the Formula

The formula

$K = \dfrac{1}{2}mv^2$

shows that kinetic energy depends on:

• the mass of the body

• the square of its speed

This means:

• if mass doubles, kinetic energy doubles

• if speed doubles, kinetic energy becomes four times

• if speed triples, kinetic energy becomes nine times

So, speed has a much greater effect on kinetic energy than mass.

Derivation of Kinetic Energy Formula

Consider a body of mass $m$ moving with initial speed $u$. A force acts on it and changes its speed to $v$ after it moves through a displacement $s$.

We know:

$W = Fs$

From Newton’s second law:

$F = ma$

So,

$W = mas$

Using the equation of motion:

$v^2 - u^2 = 2as$

or

$as = \frac{v^2 - u^2}{2}$

Substituting into the work equation:

$W = m \cdot \frac{v^2 - u^2}{2}$

$W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

This shows that work done changes the quantity $\frac{1}{2}mv^2$.

Hence, kinetic energy is defined as:

$K = \frac{1}{2}mv^2$

SI Unit and Dimensional Formula

Since kinetic energy is a form of energy, its SI unit is:

joule $(J)$

Its dimensional formula is:

$[ML^2T^{-2}]$

Important Characteristics of Kinetic Energy

• Kinetic Energy is a Scalar Quantity: It has only magnitude and no direction.

• Kinetic Energy is Never Negative: Since velocity is squared, kinetic energy is always positive or zero.

So, $K \geq 0$

• Kinetic Energy Depends on the Frame of Reference: A body may be at rest in one frame and moving in another. Therefore, its kinetic energy depends on the observer.

• For example, a passenger sitting inside a moving bus is at rest relative to the bus but moving relative to the ground.

• Kinetic Energy Depends on Speed, Not Direction: If a body moves with the same speed in opposite directions, its kinetic energy remains the same because speed is squared.

Relation Between Kinetic Energy and Momentum

Momentum of a body is:

$p = mv$

Kinetic energy is:

$K = \frac{1}{2}mv^2$

Substituting

$v = \frac{p}{m}$

we get:

$K = \frac{1}{2}m\left(\frac{p}{m}\right)^2$

$K = \frac{p^2}{2m}$

So, the relation between kinetic energy and momentum is:

$K = \frac{p^2}{2m}$

or

$p = \sqrt{2mK}$

This form is very useful in collision-based questions.

Dependence of Kinetic Energy on Mass and Velocity

1. For Constant Mass

If mass remains constant, then:

$K \propto v^2$

So,

$\frac{K_1}{K_2} = \frac{v_1^2}{v_2^2}$

2. For Constant Velocity

If velocity remains constant, then:

$K \propto m$

So,

$\frac{K_1}{K_2} = \frac{m_1}{m_2}$

3. Change in Kinetic Energy

If the speed of a body changes from $u$ to $v$, then the change in kinetic energy is:

$\Delta K = K_f - K_i$

$\Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

This expression is used directly in the work-energy theorem.

Kinetic Energy of a System of Particles

If a system contains several particles, then the total kinetic energy is the sum of the kinetic energies of all particles.

So, for particles of masses $m_1, m_2, m_3, \dots$ and speeds $v_1, v_2, v_3, \dots$:

$K_{\text{total}} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2 + \dots$

This shows that kinetic energy is additive.

Real-Life Examples of Kinetic Energy

• A moving car has kinetic energy.

• A rolling football has kinetic energy.

• Flowing water has kinetic energy.

• Wind has kinetic energy and can rotate wind turbines.

• A flying cricket ball has kinetic energy.

• A bullet has very high kinetic energy because of its high speed.

Example 1

A body of mass $4,kg$ moves with speed $3,m/s$. Find its kinetic energy.

Solution

Using

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2} \times 4 \times 3^2$

$K = 2 \times 9$

$K = 18,J$

Answer

Kinetic energy = $18,J$

Example 2

A body has momentum $10,kg,m/s$ and mass $2,kg$. Find its kinetic energy.

Solution

Using

$K = \frac{p^2}{2m}$

$K = \frac{10^2}{2 \times 2}$

$K = \frac{100}{4}$

$K = 25,J$

Answer

Kinetic energy = $25,J$

Example 3

If the speed of a body becomes twice its original value, how does its kinetic energy change?

Solution

Since

$K \propto v^2$

If speed becomes $2v$, then:

$K' = \frac{1}{2}m(2v)^2$

$K' = 4\left(\frac{1}{2}mv^2\right)$

$K' = 4K$

Answer

The kinetic energy becomes four times.

Potential Energy

Potential energy is the energy possessed by a body due to its position, arrangement, or configuration. Unlike kinetic energy, which depends on motion, potential energy is stored energy. A body may be at rest and still have potential energy if its position or state allows it to do work later.

This concept is very important because it helps explain how energy is stored and converted in systems such as falling bodies, springs, pendulums, and motion in a vertical circle.

What is Potential Energy?

Potential energy is the energy associated with the position or configuration of a body in a force field.

It is mainly defined for conservative forces, because for such forces the work done depends only on the initial and final positions and not on the path followed.

In general, the change in potential energy is related to the work done by a conservative force as:

$\Delta U = -W$

This means:

• if a conservative force does positive work, potential energy decreases

• if a conservative force does negative work, potential energy increases

Why Potential Energy is Called Stored Energy

Potential energy is called stored energy because it remains with the body or system due to its position or shape and can later convert into kinetic energy or another form of energy.

Examples:

• a stone kept at a height has gravitational potential energy

• a stretched spring has elastic potential energy

• water stored in a dam has gravitational potential energy

Gravitational Potential Energy

The most common example of potential energy in Class 11 Physics is gravitational potential energy.

When a body of mass $m$ is raised through a height $h$ against gravity near the Earth’s surface, work is done on it. This work gets stored as gravitational potential energy.

The formula is:

$U = mgh$

where:
$m$ = mass of the body
$g$ = acceleration due to gravity
$h$ = height above the reference level

Meaning of the Formula

The expression

$U = mgh$

shows that gravitational potential energy depends on:

• mass of the body

• gravitational acceleration

• height above the chosen reference point

So:

• heavier bodies have more potential energy at the same height

• higher position means greater potential energy

Reference Level in Potential Energy

Potential energy is not absolute. It is always measured relative to a chosen reference level.

Usually, we take:

$U = 0$

at ground level or any convenient point.

Then:

• above the reference level, potential energy is positive

• below the reference level, it may be taken as negative

So what really matters in Physics is the change in potential energy, not its absolute value.

Change in Gravitational Potential Energy

If a body is moved from height $h_1$ to height $h_2$, then the change in potential energy is:

$\Delta U = mg(h_2 - h_1)$

If the body moves upward, potential energy increases.

If the body moves downward, potential energy decreases.

Work Done by Gravity

When a body falls downward, gravity does positive work because force and displacement are in the same direction.

If a body falls from height $h_1$ to $h_2$, then the work done by gravity is:

$W = mg(h_1 - h_2)$

Since

$\Delta U = -W$

we get:

$\Delta U = -mg(h_1 - h_2)$

This confirms that potential energy decreases when gravity does positive work.

Elastic Potential Energy

Another important form of potential energy is the potential energy stored in a spring.

When a spring is compressed or stretched, work is done on it. This work gets stored in the spring as elastic potential energy.

For a spring of spring constant $k$ stretched or compressed by distance $x$, the elastic potential energy is:

$U = \frac{1}{2}kx^2$

where:
$k$ = spring constant
$x$ = extension or compression

Why the Formula Has $\frac{1}{2}$

The spring force does not remain constant. It increases gradually from $0$ to $kx$ as the spring is stretched or compressed. Since the force is variable, the work done is found using integration, which gives:

$U = \frac{1}{2}kx^2$

This is a very important result for numerical problems involving spring-block systems.

Relation Between Force and Potential Energy

For a conservative force, force and potential energy are related.

In one-dimensional motion:

$F = -\frac{dU}{dx}$

This means force acts in the direction in which potential energy decreases.

The negative sign indicates that conservative force always tries to bring the system toward lower potential energy.

Potential Energy Curve and Equilibrium

Potential energy diagrams are useful in understanding equilibrium.

Stable Equilibrium

A body is in stable equilibrium when, after a small displacement, it tends to return to its original position.

In stable equilibrium:

• potential energy is minimum

• $\frac{dU}{dx} = 0$

• $\frac{d^2U}{dx^2} > 0$

Example: A ball at the bottom of a bowl

Unstable Equilibrium

A body is in unstable equilibrium when, after a small displacement, it moves farther away from its original position.

In unstable equilibrium:

• potential energy is maximum

• $\frac{dU}{dx} = 0$

• $\frac{d^2U}{dx^2} < 0$

Example: A ball balanced at the top of a hill

Neutral Equilibrium

A body is in neutral equilibrium when, after a small displacement, it stays in the new position.

In neutral equilibrium:

• potential energy remains constant

Example: A ball on a horizontal surface

Potential Energy of a System

Potential energy is often a property of the system, not just of one isolated object.

For example:

• gravitational potential energy depends on the Earth-body system

• spring potential energy depends on the spring-body system

This idea is important because potential energy comes from interactions between bodies.

Real-Life Examples of Potential Energy

• A book kept on a shelf has gravitational potential energy.

• A stretched bow has elastic potential energy.

• A raised hammer has gravitational potential energy.

• A compressed spring in a toy stores elastic potential energy.

• Water stored in a dam has potential energy that can later generate electricity.

Examples of Potential Energy Example 1

A body of mass $2,kg$ is raised to a height of $5,m$. Find its gravitational potential energy.

Take $g = 9.8,m/s^2$.

Solution

Using

$U = mgh$

$U = 2 \times 9.8 \times 5$

$U = 98,J$

Answer

Gravitational potential energy = $98,J$

Examples of Potential Energy Example 2

A spring of spring constant $200,N/m$ is stretched by $0.1,m$. Find the potential energy stored in it.

Solution

Using

$U = \frac{1}{2}kx^2$

$U = \frac{1}{2} \times 200 \times (0.1)^2$

$U = 100 \times 0.01$

$U = 1,J$

Answer

Elastic potential energy = $1,J$

Examples of Potential Energy Example 3

A body moves from a height of $8,m$ to $3,m$. Find the decrease in gravitational potential energy if mass is $1,kg$ and $g = 10,m/s^2$.

Solution

Decrease in potential energy:

$\Delta U = mg(h_2 - h_1)$

$\Delta U = 1 \times 10 \times (3 - 8)$

$\Delta U = -50,J$

So, the decrease in potential energy is $50,J$.

Answer

Decrease in potential energy = $50,J$

Work-Energy Theorem

The work-energy theorem is one of the most important results in mechanics because it directly connects force and motion with energy. Instead of solving every problem using Newton’s laws and equations of motion, this theorem allows us to study motion through the change in kinetic energy of a body.

In simple words, the theorem says that when a net force acts on a body and does work on it, the kinetic energy of the body changes by exactly the same amount.

Statement of Work-Energy Theorem

The work-energy theorem states: The net work done by all the forces acting on a body is equal to the change in its kinetic energy.

Mathematically,

$W_{\text{net}} = \Delta K$

or

$W_{\text{net}} = K_f - K_i$

where:
$W_{\text{net}}$ = net work done on the body
$K_i$ = initial kinetic energy
$K_f$ = final kinetic energy

This means:

$W_{\text{net}} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where $u$ is initial speed and $v$ is final speed.

Meaning of the Theorem

This theorem tells us that:

• if net work done on a body is positive, its kinetic energy increases

• if net work done on a body is negative, its kinetic energy decreases

• if net work done is zero, the kinetic energy remains unchanged

So, the theorem gives a direct relation between work and speed.

Why the Work-Energy Theorem is Important

The work-energy theorem is useful because it often makes problem-solving easier.

Instead of finding:

• acceleration

• time

• force components in detail

we can directly relate the total work done to the change in kinetic energy.

This is especially useful in:

• variable force problems

• motion on rough surfaces

• collisions

• circular motion

• spring problems

Derivation of the Work-Energy Theorem

Consider a body of mass $m$ moving along a straight line under the action of a net force $F$. Let its initial speed be $u$ and final speed be $v$ after moving through displacement $s$.

We know that work done by the net force is:

$W_{\text{net}} = Fs$

From Newton’s second law:

$F = ma$

So,

$W_{\text{net}} = mas$

Now using the equation of motion:

$v^2 - u^2 = 2as$

or

$as = \frac{v^2 - u^2}{2}$

Substituting into the work expression:

$W_{\text{net}} = m \cdot \frac{v^2 - u^2}{2}$

$W_{\text{net}} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

But we know that:

$K_i = \frac{1}{2}mu^2$

$K_f = \frac{1}{2}mv^2$

Therefore,

$W_{\text{net}} = K_f - K_i$

Hence proved.

Work-Energy Theorem for Multiple Forces

In real situations, more than one force may act on a body at the same time. For example, a block moving on a rough surface may be acted upon by:

• applied force

• friction

• gravity

• normal reaction

The work-energy theorem still applies, but we must use the net work done by all forces.

So,

$W_{\text{applied}} + W_{\text{friction}} + W_{\text{gravity}} + W_{\text{normal}} = \Delta K$

Remember:

• normal force often does zero work if displacement is perpendicular to it

• friction usually does negative work

• gravity may do positive or negative work depending on direction of motion

Positive, Negative, and Zero Net Work

Positive Net Work

If net work is positive, then:

$K_f > K_i$

So the body speeds up.

Example: A car accelerates when engine force does more work than frictional losses.

Negative Net Work

If net work is negative, then:

$K_f < K_i$

So the body slows down.

Example: A moving ball slows down due to friction.

Zero Net Work

If net work is zero, then:

$K_f = K_i$

So the speed remains constant.

Example: In uniform circular motion, the centripetal force does zero work, so kinetic energy remains constant.

Work-Energy Theorem and Constant Force

When the force is constant, the theorem is usually straightforward to apply because:

$W = Fs\cos\theta$

Then this work is directly equated to change in kinetic energy.

So,

$Fs\cos\theta = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

This is very useful in simple numerical problems.

Work-Energy Theorem and Variable Force

Even when force is variable, the theorem remains valid.

In that case, net work is written as:

$W_{\text{net}} = \int \vec{F} \cdot d\vec{r}$

And still:

$W_{\text{net}} = \Delta K$

So the theorem is not limited to constant force only. It is a general result in mechanics.

Work-Energy Theorem and Friction

Friction is a non-conservative force and usually does negative work. Because of this, the kinetic energy of the body decreases.

Suppose a block slides on a rough horizontal surface. Then:

$W_{\text{friction}} = -fs$

Using the work-energy theorem:

$-fs = K_f - K_i$

This helps us directly find the final speed or stopping distance.

Work-Energy Theorem in Vertical Motion

In vertical motion, both gravity and any applied force may do work.

For example, if a body falls freely under gravity, then the work done by gravity is positive, so kinetic energy increases.

If a body is thrown upward, gravity does negative work, so kinetic energy decreases.

Thus, work-energy theorem is very useful in upward and downward motion problems.

Work-Energy Theorem in Circular Motion

In circular motion, the theorem can still be used. However, we must carefully identify which forces are doing work.

For example:

• centripetal force is perpendicular to displacement, so it does zero work

• gravity may do work in vertical circular motion

• tension may or may not do work depending on the situation

This idea becomes very useful in motion in a vertical circle.

Work-Energy Theorem Example 1

A force of $10,N$ acts on a body of mass $2,kg$ initially at rest and moves it through a distance of $5,m$. Find its final speed.

Solution

Work done:

$W = Fs$

$W = 10 \times 5 = 50,J$

Using the work-energy theorem:

$W = K_f - K_i$

Since the body is initially at rest, $K_i = 0$

So,

$50 = \frac{1}{2} \times 2 \times v^2$

$50 = v^2$

$v = \sqrt{50}$

$v = 5\sqrt{2},m/s$

Answer

Final speed = $5\sqrt{2},m/s$

Work-Energy Theorem Example 2

A block of mass $1,kg$ moving with speed $6,m/s$ is brought to rest by friction. Find the work done by friction.

Solution

Using the work-energy theorem:

$W = K_f - K_i$

$W = 0 - \frac{1}{2} \times 1 \times 6^2$

$W = -18,J$

Answer

Work done by friction = $-18,J$

The negative sign shows that friction removes kinetic energy from the block.

Work-Energy Theorem Example 3

A body of mass $4,kg$ moving with speed $3,m/s$ is accelerated to $5,m/s$. Find the net work done.\

Solution

Using:

$W_{\text{net}} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

$W_{\text{net}} = \frac{1}{2} \times 4 \times 5^2 - \frac{1}{2} \times 4 \times 3^2$

$W_{\text{net}} = 2(25 - 9)$

$W_{\text{net}} = 2 \times 16$

$W_{\text{net}} = 32,J$

Answer

Net work done = $32,J$

Difference Between Work-Energy Theorem and Conservation of Mechanical Energy

Work-Energy Theorem

• valid in all cases

• includes all forces

• relates net work to change in kinetic energy

Conservation of Mechanical Energy

• valid only when only conservative forces act

• says total mechanical energy remains constant

So both are related, but they are not the same concept.

Power

Power is the rate at which work is done. It tells us how quickly energy is transferred or how fast a task is completed. In many practical situations, doing the same amount of work is not enough to compare two machines or two persons. We also need to know how much time they take. That is why power becomes an important physical quantity.

For example, two students may carry the same load to the same height, so both do the same work. But if one student does it in less time, that student delivers more power.

Definition of Power

Power is defined as the work done per unit time.

Mathematically,

$P = \frac{W}{t}$

where:
$P$ = power
$W$ = work done
$t$ = time taken

This is the formula for average power when a certain amount of work is done in a finite time interval.

SI Unit of Power

The SI unit of power is watt.

$1 , \text{watt} = 1 , \text{joule per second}$

So,

$1,W = 1,J/s$

This means a device has power of $1$ watt if it does work of $1$ joule in $1$ second.

Other larger practical units are:

• kilowatt $(kW)$
  $1,kW = 1000,W$

• megawatt $(MW)$
  $1,MW = 10^6,W$

• horsepower $(hp)$
  $1,hp \approx 746,W$

Dimensional Formula of Power

Since

$P = \frac{W}{t}$

and the dimensional formula of work is:

$[ML^2T^{-2}]$

the dimensional formula of power is:

$[ML^2T^{-3}]$

Average Power

Average power is the total work done divided by the total time taken.

$P_{\text{avg}} = \frac{W}{t}$

This formula is useful when work is done uniformly or when we need the average rate of doing work over a time interval.

Example of Average Power

If a person does $200,J$ of work in $10,s$, then:

$P = \frac{200}{10} = 20,W$

So the average power is $20,W$.

Instantaneous Power

In many real situations, work may not be done at a constant rate. In such cases, we define instantaneous power.

Instantaneous power is the rate at which work is done at a particular instant.

Mathematically,

$P = \frac{dW}{dt}$

Now we know:

$dW = \vec{F} \cdot d\vec{r}$

Dividing both sides by $dt$:

$P = \frac{dW}{dt} = \vec{F} \cdot \frac{d\vec{r}}{dt}$

Since

$\frac{d\vec{r}}{dt} = \vec{v}$

we get:

$P = \vec{F} \cdot \vec{v}$

So the instantaneous power delivered by a force is:

$P = \vec{F} \cdot \vec{v}$

or

$P = Fv\cos\theta$

where $\theta$ is the angle between force and velocity.

Meaning of the Formula $P = \vec{F} \cdot \vec{v}$

This formula shows that only the component of force in the direction of velocity contributes to power.

• If force and velocity are in the same direction, power is maximum.

• If force is perpendicular to velocity, power is zero.

• If force acts opposite to velocity, power is negative.

This is similar to the concept of work, where only the component of force along displacement contributes.

Power Example 1

A machine does $500,J$ of work in $10,s$. Find its power.

Solution

Using

$P = \frac{W}{t}$

$P = \frac{500}{10}$

$P = 50,W$

Answer

Power = $50,W$

Power Example 2

A force of $20,N$ acts on a body moving with velocity $5,m/s$ in the same direction. Find the power.

Solution

Using

$P = Fv\cos\theta$

Since force and velocity are in the same direction, $\theta = 0^\circ$

$P = 20 \times 5 \times \cos 0^\circ$

$P = 100,W$

Answer

Power = $100,W$

Power Example 3

A force of $10,N$ acts perpendicular to the velocity of a body. Find the power delivered by the force.

Solution

Since force is perpendicular to velocity, $\theta = 90^\circ$

$P = Fv\cos 90^\circ$

$P = 0$

Answer

Power = $0,W$

This shows that even if force is acting, power can be zero when the force has no component along velocity.

Efficiency and Power

Power is often discussed together with efficiency in machines.

Efficiency tells us how much of the input power is converted into useful output power.

Mathematically,

$\eta = \frac{\text{output power}}{\text{input power}}$

If expressed in percentage:

$\eta = \frac{\text{output power}}{\text{input power}} \times 100$

This is useful in machines like engines, motors, and pumps.

Potential Energy of a Spring

The potential energy of a spring is the energy stored in it when it is stretched or compressed from its natural length. This energy is called elastic potential energy because it is stored due to the elastic property of the spring.

This topic is very important in Work, Energy and Power because it is one of the most common examples of energy stored due to configuration. It is also closely connected with variable force, Hooke’s law, and conservation of mechanical energy.

What is Spring Potential Energy?

When a spring is stretched or compressed, an external force has to do work against the restoring force of the spring. This work does not disappear. It gets stored in the spring as potential energy.

So, the potential energy of a spring is the energy stored in the spring because of its deformation.

Hooke’s Law

For a light spring within the elastic limit, the restoring force is directly proportional to the displacement from its natural length.

Mathematically,

$F = -kx$

where:
$F$ = restoring force of the spring
$k$ = spring constant
$x$ = displacement from natural length

The negative sign shows that the restoring force always acts opposite to the displacement.

• If the spring is stretched, the spring force acts backward.

• If the spring is compressed, the spring force acts forward toward the mean position.

Why Spring Force is a Variable Force

The spring force is not constant. As the extension or compression increases, the magnitude of the force also increases.

• At $x = 0$, force is zero

• At larger $x$, force becomes larger

That is why the work done in stretching or compressing a spring cannot be found using the simple formula for constant force. We need integration.

Derivation of Potential Energy of a Spring

Suppose a spring is stretched from its natural length by a distance $x$.

The external force needed at a small extension $x$ is:

$F = kx$

The small work done in stretching it further by a small distance $dx$ is:

$dW = F,dx$

$dW = kx,dx$

Total work done in stretching the spring from $x = 0$ to $x$ is:

$W = \int_0^x kx,dx$

$W = k \int_0^x x,dx$

$W = k \left[\frac{x^2}{2}\right]_0^x$

$W = \frac{1}{2}kx^2$

This work done on the spring is stored in it as elastic potential energy.

Therefore, the potential energy stored in a spring is:

$U = \frac{1}{2}kx^2$

Meaning of the Formula

The formula

$U = \frac{1}{2}kx^2$

shows that spring potential energy depends on:

• spring constant $k$

• square of deformation $x$

This means:

• a stiffer spring stores more energy for the same extension

• if extension doubles, potential energy becomes four times

So spring potential energy increases rapidly with displacement.

Work Done by Spring and Work Done on Spring

This is one of the most important distinctions in spring problems.

Work Done on the Spring

When an external force stretches or compresses the spring, the work done on the spring is positive because energy is being stored in it.

$W_{\text{on spring}} = \frac{1}{2}kx^2$

Work Done by the Spring

When the spring returns toward its natural length, it does work on the attached object. In that case, the work done by the spring is negative of the work done on it.

$W_{\text{by spring}} = -\frac{1}{2}kx^2$

More generally, if the spring moves from deformation $x_1$ to $x_2$, then the work done by the spring is:

$W = \frac{1}{2}k(x_1^2 - x_2^2)$

This formula is very useful in numerical problems.

Spring Potential Energy in Stretching and Compression

The formula

$U = \frac{1}{2}kx^2$

is valid for both stretching and compression.

This is because $x^2$ is always positive.

So:

• a spring stretched by $x$ stores the same energy as a spring compressed by the same amount

Only the direction of the spring force changes, not the stored energy formula.

Force-Displacement Graph for a Spring

If spring force is plotted against displacement, the graph is a straight line through the origin.

Since

$F = kx$

the force increases linearly with displacement.

The work done in stretching the spring from $0$ to $x$ is equal to the area under the force-displacement graph.

That area is a triangle:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

$= \frac{1}{2} \times x \times kx$

$= \frac{1}{2}kx^2$

This gives the same expression for spring potential energy.

Relation Between Spring Force and Potential Energy

For a conservative force, force is related to potential energy by:

$F = -\frac{dU}{dx}$

For spring potential energy,

$U = \frac{1}{2}kx^2$

Differentiating,

$\frac{dU}{dx} = kx$

So,

$F = -kx$

This confirms Hooke’s law and shows that spring force is conservative.

Spring as a Conservative System

Spring force is a conservative force. This means:

• work done by spring depends only on initial and final positions

• work done over a closed path is zero

• energy stored in the spring can be fully recovered if there is no friction

This is why spring systems are often used in conservation of mechanical energy problems.

Spring-Block System and Energy Conversion

In a spring-block system, energy keeps changing form:

• at maximum extension or compression, kinetic energy is zero and spring potential energy is maximum

• at mean position, spring potential energy is zero and kinetic energy is maximum

This continuous exchange of energy is a very important application of spring potential energy.

Real-Life Examples of Spring Potential Energy

• a stretched slingshot stores elastic potential energy

• a compressed spring in a toy car stores energy

• a bow before releasing an arrow stores elastic potential energy

• a spring mattress stores energy when compressed

• shock absorbers and spring mechanisms in machines also work on this principle