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Work, Energy and Power

Master Work, Energy and Power for JEE Main, JEE Advanced, NEET & Board Exams. Covers work-energy theorem, kinetic & potential energy, conservation of energy, power, elastic & inelastic collisions, and potential energy curves with solved examples.

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Work, Energy and Power are among the most fundamental concepts in physics, forming a bridge between kinematics and dynamics. The work-energy theorem and the principle of conservation of energy are powerful tools that often let us solve problems — especially those involving variable forces — far more elegantly than applying Newton's laws directly. For JEE (Main & Advanced), NEET, and board exams, this chapter is extremely high-weightage, with questions ranging from straightforward formula application to multi-concept problems involving springs, collisions, and circular motion.

1. Work Done by a Force

Work is done when a force causes (or tends to cause) displacement of a body. It is a scalar quantity defined as:

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

where $θ$ is the angle between the force $\vec{F}$ and displacement $\vec{s}$ . SI unit of work is the Joule (J): $1 J = 1 N·m$ .

Angle $θ$	Work Done	Example
$0 °$	$W = F s$ (maximum, positive)	Pushing a box in direction of motion
$90 °$	$W = 0$	Normal force, centripetal force
$180 °$	$W = - F s$ (maximum negative)	Friction opposing motion
$0 ° < θ < 90 °$	$W > 0$ (positive)	Force at acute angle to motion
$90 ° < θ < 180 °$	$W < 0$ (negative)	Force opposing but not fully reversed

Work Done by a Variable Force

When force is not constant, work is calculated by integration:

$W = \int_{x_{i}}^{x_{f}} F (x) d x$

Graphically, work done = area under the $F$ – $x$ graph (with sign). Area above the $x$ -axis is positive work; area below is negative work.

Work Done Against Gravity

When a body of mass $m$ is raised through height $h$ (regardless of the path taken):

$W_{g r a v i t y} = - m g h$ (work done BY gravity)

$W_{a g a i n s t g r a v i t y} = + m g h$ (work done AGAINST gravity)

Gravity does negative work when the body moves up and positive work when the body moves down.

2. Kinetic Energy and the Work-Energy Theorem

Kinetic Energy (KE)

The energy possessed by a body by virtue of its motion is:

$K E = \frac{1}{2} m v^{2} = \frac{p^{2}}{2 m}$

where $p = m v$ is the linear momentum. KE is always non-negative. SI unit: Joule (J).

Work-Energy Theorem

The net work done by all forces acting on a body equals the change in its kinetic energy:

$W_{n e t} = Δ K E = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$

This theorem holds for both constant and variable forces.
It is a scalar equation — much easier to apply than Newton's vector equations when forces are complex.
If $W_{n e t} > 0$ : KE increases (body speeds up). If $W_{n e t} < 0$ : KE decreases (body slows down). If $W_{n e t} = 0$ : KE constant (speed unchanged).

Relation Between KE and Momentum

$K E = \frac{p^{2}}{2 m}$ $\Rightarrow$ $p = \sqrt{2 m \cdot K E}$

Change	Effect on KE	Effect on $p$
Mass doubled, speed same	KE doubled	$p$ doubled
Speed doubled, mass same	KE quadrupled	$p$ doubled
Same KE, different masses	—	Heavier body has more $p$
Same $p$ , different masses	Lighter body has more KE	—

3. Potential Energy

Potential energy (PE) is the energy stored in a body by virtue of its position or configuration. It is associated only with conservative forces.

(i) Gravitational Potential Energy

$U_{g} = m g h$

Here $h$ is the height above a chosen reference level (usually ground). PE depends only on height — the path taken does not matter (conservative force).

(ii) Elastic Potential Energy (Spring)

When a spring of spring constant $k$ is compressed or stretched by $x$ from its natural length:

$U_{s p r i n g} = \frac{1}{2} k x^{2}$

Work done by spring force when stretched from $x_{1}$ to $x_{2}$ :

$W_{s p r i n g} = - (\frac{1}{2} k x_{2}^{2} - \frac{1}{2} k x_{1}^{2}) = - Δ U_{s p r i n g}$

Relation Between Conservative Force and Potential Energy

$F = - \frac{d U}{d x}$

A conservative force always acts in the direction of decreasing potential energy — from higher PE to lower PE.

Conservative vs. Non-Conservative Forces

Conservative Forces	Non-Conservative Forces
Work done is path-independent	Work done depends on path
Work done in a closed loop = 0	Work done in a closed loop ≠ 0
Potential energy can be defined	Potential energy cannot be defined
Examples: Gravity, spring, electrostatic	Examples: Friction, air drag, viscosity

4. Conservation of Mechanical Energy

Mechanical Energy is the sum of kinetic and potential energy: $E = K E + P E$ .

For a system acted upon only by conservative forces, total mechanical energy is conserved:

$K E_{i} + P E_{i} = K E_{f} + P E_{f} = constant$

i.e., $\frac{1}{2} m v_{i}^{2} + U_{i} = \frac{1}{2} m v_{f}^{2} + U_{f}$

When non-conservative forces (e.g., friction) act, energy is not conserved — mechanical energy decreases:

$W_{n o n - c o n s e r v a t i v e} = Δ E_{m e c h a n i c a l}$
$= (K E_{f} + P E_{f}) - (K E_{i} + P E_{i})$

The lost mechanical energy is converted into heat, sound, or other forms — total energy is still conserved.

Application: Velocity at Any Height During Free Fall / Projectile

Using energy conservation for a body falling from height $H$ to height $h$ (smooth surface):

$v = \sqrt{v_{0}^{2} + 2 g (H - h)}$

At the bottom ( $h = 0$ ): $v = \sqrt{v_{0}^{2} + 2 g H}$ . If released from rest: $v = \sqrt{2 g H}$ .

Application: Block-Spring System

When a block of mass $m$ moving with speed $v$ compresses a spring of constant $k$ by maximum $x_{m a x}$ :

$\frac{1}{2} m v^{2} = \frac{1}{2} k x_{m a x}^{2} ⟹ x_{m a x} = v \sqrt{\frac{m}{k}}$

5. Power

Power is the rate of doing work or the rate of energy transfer:

$P = \frac{d W}{d t} = \vec{F} \cdot \vec{v} = F v \cos θ$

SI unit: Watt (W): $1 W = 1 J/s$ . Other units: $1 hp (horsepower) = 746 W$ ; $1 kWh = 3.6 \times 10^{6} J$ .

Power is a scalar quantity.
For a constant force: $P_{a v g} = \frac{W}{t} = \frac{F s \cos θ}{t} = F v_{a v g} \cos θ$
Instantaneous power: $P = F v$ (when $\vec{F} ∥ \vec{v}$ )
At maximum speed on a level road: driving force = friction (acceleration = 0), so $P = f_{k} \cdot v_{m a x}$ .

Important Result: Vehicle on a Road

A vehicle engine of power $P$ moving against resistance $f$ has maximum speed:

$v_{m a x} = \frac{P}{f}$

On an inclined road with angle $θ$ , total resistance = $f + m g \sin θ$ , so $v_{m a x} = \frac{P}{f + m g \sin θ}$ .

6. Collisions

In all collisions, linear momentum is conserved (if no external forces act). Kinetic energy may or may not be conserved.

Type	Momentum	KE	$e$	Example
Perfectly Elastic	Conserved	Conserved	$e = 1$	Billiard balls (ideal)
Perfectly Inelastic	Conserved	Not conserved (max loss)	$e = 0$	Clay/putty collision
Inelastic	Conserved	Partially lost	$0 < e < 1$	Most real collisions

Coefficient of Restitution ( $e$ )

$e = \frac{Relative speed of separation}{Relative speed of approach} = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$

Head-On Elastic Collision between $m_{1}$ (speed $u_{1}$ ) and $m_{2}$ (at rest)

$v_{1} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} u_{1}$ $v_{2} = \frac{2 m_{1}}{m_{1} + m_{2}} u_{1}$

Special Cases of Elastic Collision

Condition	$v_{1}$	$v_{2}$	Inference
$m_{1} = m_{2}$	$0$	$u_{1}$	Velocities exchanged
$m_{1} ≫ m_{2}$	$\approx u_{1}$	$\approx 2 u_{1}$	Heavy hits light — light bounces at $2 u_{1}$
$m_{1} ≪ m_{2}$	$\approx - u_{1}$	$\approx 0$	Light hits heavy — bounces back at same speed

Perfectly Inelastic Collision — Common Velocity and Energy Loss

After collision, both bodies move together with common velocity:

$v_{c o m m o n} = \frac{m_{1} u_{1} + m_{2} u_{2}}{m_{1} + m_{2}}$

Loss in KE:

$Δ K E = \frac{1}{2} \cdot \frac{m_{1} m_{2}}{m_{1} + m_{2}} (u_{1} - u_{2})^{2}$

Oblique Elastic Collision

For an oblique collision, resolve velocities along and perpendicular to the line of impact. Along the line of impact, the collision equations apply; perpendicular to it, each body's velocity component remains unchanged (no impulse perpendicular to line of impact for smooth spheres).

7. Potential Energy Curve and Equilibrium

A graph of potential energy $U (x)$ vs. position $x$ reveals the nature of equilibrium of a system. Since $F = - \frac{d U}{d x}$ :

Stable Equilibrium: At a minimum of $U$ . $\frac{d U}{d x} = 0$ and $\frac{d^{2} U}{d x^{2}} > 0$ . A small displacement brings the body back to equilibrium.
Unstable Equilibrium: At a maximum of $U$ . $\frac{d U}{d x} = 0$ and $\frac{d^{2} U}{d x^{2}} < 0$ . A small displacement causes the body to move further away.
Neutral Equilibrium: $U$ is constant over a region. $\frac{d U}{d x} = 0$ everywhere in the region. Body remains wherever placed.

The body cannot exist in a region where $U > E_{t o t a l}$ (since KE would become negative, which is physically impossible). The points where $U = E_{t o t a l}$ are called turning points.

Energy Method vs. Newton's Law — When to Use Which?

Use the Work-Energy Theorem when the problem asks for speed/velocity at a point and involves variable forces or complex paths — especially spring forces, curved tracks, and inclined planes. You avoid dealing with vector components at every instant.
Use Newton's Laws (FBD) when the problem asks for forces (tension, normal reaction) at a specific point, or when acceleration is required at an instant. For example, finding normal force at the bottom of a circular loop requires Newton's Second Law, not just energy conservation.
The most powerful approach for JEE Advanced problems is often to combine both: use energy conservation to find speed at a point, then apply Newton's Second Law at that point to find forces.

JEE/NEET Power Tips — Avoid These Common Mistakes

Work done by normal force and centripetal force is always zero — both are perpendicular to displacement/velocity at every instant.
KE is always ≥ 0. If energy conservation gives a negative KE at some point, the body cannot reach that point — it turns back earlier (check the turning point).
In a perfectly inelastic collision, KE is NOT zero — only the maximum possible KE is lost. Some KE remains as long as the combined body is moving.
$1 kWh$ is a unit of energy, not power. $1 kWh = 3.6 \times 10^{6} J$ . Electricity bills are in kWh.
Coefficient of restitution $e$ is between 0 and 1 for real collisions. $e > 1$ is physically impossible for ordinary collisions (would imply generation of energy).
Conservation of energy applies to the entire system, not individual bodies. Always identify the system and account for all energy forms (KE, PE, heat due to friction).
Work done by friction on a system = $- μ_{k} N \times$ (relative sliding distance), NOT just $μ_{k} N \times$ displacement of one body.

Practice Questions (JEE / NEET / Board Level)

Q1: A body of mass $2$ kg is thrown vertically upward with a speed of $10$ m/s. What is its kinetic energy when it has risen to a height of $3$ m? ( $g = 10$ m/s²)

A) $100$ J
B) $60$ J
C) $40$ J
D) $0$ J

Answer: C) $40$ J.
Using energy conservation:
$K E = K E_{i} - m g h$
$= \frac{1}{2} (2) (10)^{2} - (2) (10) (3)$
$= 100 - 60 = 40$ J.

Q2: A particle moves under a force $F = (3 x^{2} + 2 x)$ N along the $x$ -axis. Work done by the force as the particle moves from $x = 0$ to $x = 2$ m is:

A) $12$ J
B) $8$ J
C) $16$ J
D) $14$ J

Answer: A) $12$ J.
$W = \int_{0}^{2} (3 x^{2} + 2 x) d x$
$= {[x^{3} + x^{2}]}_{0}^{2} = (8 + 4) - 0 = 12$ J.
Answer: A) 12 J.

Q3: A ball of mass $0.1$ kg is dropped from a height of $5$ m and bounces back to a height of $3.2$ m. The coefficient of restitution is:

A) $0.64$
B) $0.8$
C) $0.36$
D) $0.5$

Answer: B) $0.8$ . Speed just before hitting ground: $u = \sqrt{2 g \times 5} = \sqrt{100} = 10$ m/s. Speed just after bouncing: $v = \sqrt{2 g \times 3.2} = \sqrt{64} = 8$ m/s. $e = \frac{v}{u} = \frac{8}{10} = 0.8$ .

Q4: A body of mass $4$ kg moving with velocity $6$ m/s collides head-on and perfectly inelastically with a body of mass $2$ kg at rest. The loss in kinetic energy is:

A) $48$ J
B) $24$ J
C) $72$ J
D) $12$ J

Answer: B) $24$ J.
$Δ K E = \frac{1}{2} \cdot \frac{m_{1} m_{2}}{m_{1} + m_{2}} (u_{1} - u_{2})^{2}$
$= \frac{1}{2} \cdot \frac{4 \times 2}{4 + 2} (6 - 0)^{2}$
$= \frac{1}{2} \cdot \frac{8}{6} \cdot 36$
$= \frac{1}{2} \times \frac{4}{3} \times 36 = 24$ J.

Q5: An engine of power $3$ kW pulls a train of mass $10^{4}$ kg on a level track. If the frictional force is $500$ N, the maximum speed of the train is:

A) $3$ m/s
B) $6$ m/s
C) $9$ m/s
D) $12$ m/s

Answer: B) $6$ m/s. At maximum speed, driving force = friction. $v_{m a x} = \frac{P}{f} = \frac{3000}{500} = 6$ m/s.

Q6 (JEE Advanced type): A block of mass $m$ is released from rest on a smooth incline of angle $θ$ at a height $h$ above the ground. At the bottom of the incline it hits a spring of spring constant $k$ . The maximum compression of the spring is:

A) $\sqrt{\frac{2 m g h}{k}}$
B) $\frac{m g h}{k}$
C) $\sqrt{\frac{m g \sin θ}{k}}$
D) $\frac{2 m g h}{k}$

Answer: A) $\sqrt{\frac{2 m g h}{k}}$ . At maximum compression $x$ , the block is momentarily at rest. Energy conservation (taking bottom as reference, with spring compressed by $x$ so block is at height $0$ effectively on a smooth surface): $m g h = \frac{1}{2} k x^{2} ⟹ x = \sqrt{\frac{2 m g h}{k}}$ .

Q7 (JEE Advanced type): The potential energy of a particle is given by $U (x) = x^{2} - 4 x + 3$ (in joules, $x$ in metres). The particle is in stable equilibrium at:

A) $x = 0$ m
B) $x = 2$ m
C) $x = 4$ m
D) $x = 3$ m

Answer: B) $x = 2$ m. For equilibrium: $\frac{d U}{d x} = 2 x - 4 = 0 ⟹ x = 2$ m. Since $\frac{d^{2} U}{d x^{2}} = 2 > 0$ , this is a minimum of $U$ — hence stable equilibrium.

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Work, Energy and Power

1. Work Done by a Force

Work Done by a Variable Force

Work Done Against Gravity

2. Kinetic Energy and the Work-Energy Theorem

Kinetic Energy (KE)

Work-Energy Theorem

Relation Between KE and Momentum

3. Potential Energy

(i) Gravitational Potential Energy

(ii) Elastic Potential Energy (Spring)

Relation Between Conservative Force and Potential Energy

Conservative vs. Non-Conservative Forces

4. Conservation of Mechanical Energy

Application: Velocity at Any Height During Free Fall / Projectile

Application: Block-Spring System

5. Power

Important Result: Vehicle on a Road

6. Collisions

Coefficient of Restitution ( $e$ )

Head-On Elastic Collision between $m_{1}$ (speed $u_{1}$ ) and $m_{2}$ (at rest)

Special Cases of Elastic Collision

Perfectly Inelastic Collision — Common Velocity and Energy Loss

Oblique Elastic Collision

7. Potential Energy Curve and Equilibrium

Energy Method vs. Newton's Law — When to Use Which?

Practice Questions (JEE / NEET / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

We Value Your Privacy

Edvaya Target

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Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Work, Energy and Power

1. Work Done by a Force

Work Done by a Variable Force

Work Done Against Gravity

2. Kinetic Energy and the Work-Energy Theorem

Kinetic Energy (KE)

Work-Energy Theorem

Relation Between KE and Momentum

3. Potential Energy

(i) Gravitational Potential Energy

(ii) Elastic Potential Energy (Spring)

Relation Between Conservative Force and Potential Energy

Conservative vs. Non-Conservative Forces

4. Conservation of Mechanical Energy

Application: Velocity at Any Height During Free Fall / Projectile

Application: Block-Spring System

5. Power

Important Result: Vehicle on a Road

6. Collisions

Coefficient of Restitution (e)

Head-On Elastic Collision between m1 (speed u1) and m2 (at rest)

Special Cases of Elastic Collision

Perfectly Inelastic Collision — Common Velocity and Energy Loss

Oblique Elastic Collision

7. Potential Energy Curve and Equilibrium

Energy Method vs. Newton's Law — When to Use Which?

Practice Questions (JEE / NEET / Board Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

We Value Your Privacy

Coefficient of Restitution ( $e$ )

Head-On Elastic Collision between $m_{1}$ (speed $u_{1}$ ) and $m_{2}$ (at rest)