WORK, ENERGY AND POWER

This chapter discusses the concepts of work and energy, highlighting their
relationship and the importance of the law of energy conservation in understanding
motion and natural phenomena. When the forces involved in a process are unknown
or complex, this law becomes an essential tool for solving problems.

Work
In physics, work is done by a force when it acts on a body and causes displacement
of the point of application in the direction of the force.

For example, if a constant force F displaces a body through a distance s, the work
done W is given by:

→ →
𝑊 = 𝐹𝑠𝑐𝑜𝑠⁡θ = 𝐹. 𝑠

Where, 𝑠 is the magnitude of displacement and θ is the angle between force and
displacement. The SI unit of work is the Joule (J), which is equivalent to a
Newton-meter (N·m).
Sign Convention of Work

◦
1.​ Positive Work: When 0 < θ < 90 , the angle between the force and
displacement is acute, and the work done is positive. This occurs when the force
supports the motion of the body, such as when a force is applied in the
direction of the body’s movement.

​ 𝑊 = 𝐹𝑠𝑐𝑜𝑠⁡θ (is positive)

◦ ◦
2.​ Negative Work: When 90 < θ < 180 , the angle between the force and
displacement is obtuse, and the work done is negative. This occurs when the
force opposes the motion of the body, such as in the case of friction or resistive
forces.​

​ 𝑊 = 𝐹𝑠𝑐𝑜𝑠⁡θ (is negative)

Nature of Work

→ →
Work done is represented by the equation: 𝐹. 𝑆

Based on this, three possible cases for the nature of work done are:

1. Positive Work: If the angle between the force and displacement vectors is acute
(less than 90°), the work is positive. For example, when a horse pulls a cart on a level
road, the work done by the horse is positive.

2. Zero Work: If the force and displacement vectors are perpendicular (90° angle), no
work is done. For example, when a body moves along a circular path, the work done
by the string is zero because the force is always perpendicular to the displacement.
3. Negative Work: If the angle between the force and displacement vectors is obtuse
(greater than 90°), the work is negative. An example is when a body slides over a
rough surface, and the work done by the frictional force is negative due to the angle
being 180°.

Additionally, zero work is done in the following scenarios:

a)​ If the point of application of the force does not change, but the body moves.

b)​ If the body does not move, but the point of application of the force moves.

Work Done by A Variable Force

When force is expressed as an arbitrary function of position, calculus principles are

essential to determine the work done.

Figure below illustrates F(x) as a function of x. To proceed with our calculations, we

approximate the varying force by dividing it into several small steps. In Figure below,
the area under each segment of the curve is roughly represented by the area of a
rectangle. For each rectangle, the work done is represented by ∆𝑊𝑛 = 𝐹𝑛∆𝑥𝑛, where Fn​

is the force over a small displacement Δxn. Thus, the total work done can be
approximated by summing the areas of these rectangles: 𝑊≈∑𝐹𝑛∆𝑥𝑛​.

As the size of each step Δx decreases, the tops of the rectangles more closely align
with the actual curve depicted in the figure. In the limit, as Δx→0, which implies an
infinite number of steps, this discrete summation becomes a continuous integral.
Therefore, the work done W is given by:

𝑊 = ∫𝐹(𝑥) 𝑑𝑥

m is the mass of the object,

µ is the coefficient of friction,
2
( )
g is the acceleration due to gravity 9. 8 𝑚/𝑠 ,

I is the distance traveled by the block along the rough surface.

Similarly, work done along an inclined surface with an angle θ from horizontal is given
by µ𝑚𝑔𝑙𝑐𝑜𝑠⁡θ

Conservative and Non-Conservative Forces

A force is categorized as conservative if the work it does in moving a particle
between two points depends only on the initial and final positions, not on the path
taken. In other words, for a conservative force, the work done around a closed path is
zero.

Examples of conservative forces include gravitational, electric, and spring forces.

Typically, all central forces (those directed toward a center point) are conservative.

In contrast, a non-conservative force is one where the work done in moving a

particle from one point to another depends on the specific path taken. For
non-conservative forces, the work done around a closed path is not zero.

Examples of non-conservative forces include frictional and viscous forces; as they

oppose motion, energy is lost (e.g., as heat or sound) when overcoming these forces.
Thus, they lead to irreversible energy transformations and are categorized as
non-conservative.
Work Done Against Friction

Frictional force always acts opposite to the direction of motion (and displacement),
making the work done by friction negative. Energy expended against friction is often
dissipated as heat or sound, further classifying friction as a non-conservative force.

The work done by friction is either negative or zero, but it is never positive. Friction
always resists motion along a surface, whether horizontal or inclined:

1.​ Work done on a horizontal surface = μ𝑚𝑔𝑙,

where, m is the mass of the object, μ is the coefficient of friction, g is the
acceleration due to gravity (9.8 m/s2), l is the distance traveled along the rough
surface.

2.​ Work done on an inclined surface (at angle θ to the horizontal) = − μ𝑚𝑔𝑙𝑐𝑜𝑠θ.

Power

Power is defined as the rate at which work is done. If an amount of work ΔW is

completed in a time interval Δt, then the average power Pn is expressed as
𝑃𝑛 = Δ𝑊Δ𝑡𝑃𝑛​. Instantaneous power, on the other hand, is given by:

Δ𝑊 𝑑𝑊
𝑃 = 𝑙𝑖𝑚Δ𝑡→0 = Δ𝑡
= 𝑑𝑡
When a force F acts on an object causing an infinitesimally small displacement
dsdsds, the work done 𝑑𝑊 = 𝐹⋅𝑑𝑠. Hence, instantaneous power can be calculated as:

𝑑𝑊
𝑃= 𝑑𝑡
= 𝐹⋅𝑣
where F is the force and v is the velocity. The SI unit of power is the watt (W), or joule
per second (J/s), and power is a scalar quantity with dimensions M1L2T−3 .
Energy

Energy signifies a body's capacity to perform work. As a scalar quantity, it shares the
same unit as work (joules in SI units). In mechanics, energy is represented mainly
through kinetic and potential energy, both influencing the body's motion and state.

Potential Energy

Potential energy is the energy that a body possesses due to its position or state. This
form of energy is independent of the method used to reach the current state and
relies on the reference level chosen.

The change in potential energy (ΔU) of a system is defined as the negative of the
work done by the conservative force acting within it:

Δ𝑈 = − 𝑊𝐴𝐵​
where WAB​ is the work done by conservative forces from the initial state A to the final
state B. Since potential energy only depends on conservative forces, it cannot be
defined for non-conservative forces, as their work depends on the path taken.

Gravitational Potential Energy (GPE)

Lifting an object to height h does work against gravity, storing energy as

gravitational potential energy. The work done is 𝑚𝑔ℎ, where m is the object's mass
and g is gravitational acceleration. This also represents the increase in potential
energy. Lowering the object by height h decreases the potential energy by 𝑚𝑔ℎ.

Elastic Potential Energy

When a spring is compressed or stretched, work is performed against its restoring

force, storing energy in the spring as elastic potential energy.
Nature of Restoring Force

When a spring is displaced by a distance x (either extended or compressed), it exerts

a restoring force to counteract this displacement.

For a spring, the natural (or unstretched) length serves as the reference point, where
the potential energy is universally considered zero. This assumption standardizes
calculations across different contexts involving spring systems.

For Stretching
𝑓→ → 𝑥 1 2
𝑈𝑓 − 𝑈𝑖 =− ∫𝑖 𝐹. 𝑑𝑆; 𝑈𝑓 − 0 =− ∫0𝑖 𝑘𝑥(− 𝑖)(𝑑𝑥)𝑖; 𝑈 = 2
𝑘𝑥1

For Compression
𝑓→ → 𝑥 1 2
𝑈𝑓 − 𝑈𝑖 =− ∫𝑖 𝐹 · 𝑑𝑆 =− ∫0𝑖 𝑘𝑥𝑖(𝑑𝑥)(− 𝑖) = 𝑈 = 2
𝑘𝑥

Thus, if the spring is either stretched or compressed from natural length by x the
1 2
corresponding potential energy is 2
𝑘𝑥
Relationship between Force and Potential Energy
Now, let's explore the relationship between force and potential energy

.
Assume a body is moved from point A to point B in such a way that there is no overall
change in its kinetic energy. This means that all of the work done by the external
forces is used solely to alter the potential energy of the system, without affecting its
kinetic energy.

⇒ Work done =− change in P.E.;

⇒ 𝐹∆𝑟 = 𝑈 − (𝑈 + Δ𝑈) =− Δ𝑈
⇒ 𝐹𝑎𝑣𝑔 =− ( ) if ∆𝑟→0; 𝐹 =− 𝑙𝑖𝑚
∆𝑈
∆𝑟
∆𝑢
∆𝑟→0 ∆𝑟
=−
∂𝑈
∂𝑟

Kinetic Energy
Kinetic energy (KE) is the energy possessed by a body due to its motion. Therefore,
an object of mass m moving at a velocity v has a kinetic energy

1 2
𝐾= 2
𝑚𝑣
Since velocity is a relative quantity, kinetic energy is also a relative property.

Important Points

1.​ Since mass (m) and the square of velocity (v2) are always positive, kinetic
energy is always positive and does not depend on the direction of the motion.
2.​ Kinetic energy depends on the frame of reference. For instance, for a person of
mass m in a moving train with speed v, the kinetic energy is zero in the frame
2
of reference of the train but is 1
2
𝑚𝑣 in the frame of reference of the Earth.
Note: Energy cannot be negative. While kinetic energy is always positive, the total
energy (kinetic + potential) can be negative depending on the reference point chosen
for potential energy.

Equilibrium

In the context of "Laws of Motion," a body is said to be in translational equilibrium if

→
the net force acting on it is zero, that is, 𝐹𝑛𝑒𝑡 = 0 . If the acting forces are
𝑑𝑈
conservative, we use the relation 𝐹 = − 𝑑𝑟
for equilibrium:

→ 𝑑𝑈 𝑑𝑈
𝐹 = 0→ − 𝑑𝑟
= 0→ 𝑑𝑟
= 0
This implies that at the equilibrium point, the slope of the potential energy ( U ) versus
position ( r ) graph is zero, meaning the potential energy is at an optimum (maximum,
minimum, or constant). There are three types of equilibrium: stable, unstable, and
neutral. These conditions for equilibrium are defined as:

2
𝑑𝑈
(a) If 2 > 0 , it indicates a stable equilibrium.
𝑑𝑟

2
𝑑𝑈
(b) If 2 < 0 , it indicates an unstable equilibrium.
𝑑𝑟

2
𝑑𝑈
(c) If 2 = 0, it indicates a neutral equilibrium.
𝑑𝑟

A system naturally seeks to minimize its energy. The types of equilibrium can be
categorized on this basis:

1.​ Stable Equilibrium: If the system is slightly disturbed, it tends to return to its
original position, indicating a state of minimum potential energy.
2.​ Unstable Equilibrium: A slight disturbance causes the system to move to a
different configuration, indicating a state of maximum potential energy.
3.​ Neutral Equilibrium: The potential energy is constant, and any small
disturbance results in a new equilibrium state without further changes.
Work-Energy Theorem

When a particle is acted upon by various forces and undergoes a displacement, its
kinetic energy changes by an amount equal to the net work (Wnet) done on it by all
forces:

𝑊𝑛𝑒𝑡 = 𝐾𝑓 − 𝐾𝑖 = ∆𝐾
This equation is referred to as the work-energy theorem. It holds true regardless of
whether the forces are constant or varying, or whether the path is straight or curved.

Expanding further, we can write:

𝑊𝑐 + 𝑊𝑁𝐶 + 𝑊0𝑡ℎ = ∆𝐾
Where, Wc is the work done by conservative forces, WNC is the work done by
non-conservative forces, WOth is the work done by other forces not categorized as
conservative, non-conservative, or pseudo-forces.

Since 𝑊𝑐 = ∆𝑈 (based on the definition of potential energy), the above expression

can be modified as:

𝑊𝑁𝐶 + 𝑊𝑂𝑡ℎ = ∆𝐾 + ∆𝑈 + ∆(𝐾 + 𝑈) = ∆𝐸

In this equation, K + U = E represents the mechanical energy of the system.

When frictional forces are present, energy is dissipated in an amount equal to the
work done by friction. The energy lost through friction is converted into heat. In
practical scenarios, machine operators take various steps to minimize friction and
reduce energy loss, often using lubricants and rollers to optimize efficiency.
Derivation of Work-Energy Theorem

Let us now derive a relationship between the work done on a particle and its change
in speed. Referring to Figure, we observe that a particle moves from point P1 to point
→
P2 under the influence of a net force 𝐹 :

→ →
𝑊 = ∫𝐹⋅𝑑 𝑟
Considering the force components,

→ ^ ^ ^
𝐹 = 𝐹𝑥 𝑖 + 𝐹𝑦 𝑗 + 𝐹𝑧 𝑘 and

→ ^ ^ ^
𝑑𝑟 = 𝑑𝑥 𝑖 + 𝑑𝑦 𝑗 + 𝑑𝑧 𝑘

The work done becomes:

𝑊 = ∫(𝐹𝑥𝑑𝑥 + 𝐹𝑦𝑑𝑦 + 𝐹𝑧𝑑𝑧)

It is evident that if a particle moves along a curved path from point P1 to P2 , it does
so under the influence of a force F that varies in both direction and magnitude. The
component Fx is related to the acceleration in the x-direction:

𝑑𝑣𝑥
𝐹𝑥 = 𝑚 𝑎𝑥 = 𝑚 𝑑𝑡

Integrating both sides, we have:

𝑑𝑣𝑥
∫ 𝐹𝑥𝑑𝑥 = ∫𝑚 𝑑𝑡
𝑑𝑥

Treating vx as a function of position, we use the relation:

𝑑𝑣𝑥 𝑑𝑣𝑥 𝑑𝑥 𝑑𝑣𝑥

𝑑𝑡
= 𝑑𝑥
⋅ 𝑑𝑡
= 𝑣𝑥 𝑑𝑥

Thus:
 𝑑𝑣𝑥
∫𝐹𝑥𝑑𝑥 = ∫𝑚 𝑣𝑥 𝑑𝑥
𝑑𝑥 = ∫ 𝑚𝑣𝑥𝑑𝑣𝑥

After integration:

1 2 1 2
2
𝑚 𝑣𝑥 − 2
𝑚 𝑣𝑥
2 1

Where 𝑣𝑥 and 𝑣𝑥 are the velocities in the x-direction at points P1 and P2, respectively.
1 2

Applying the same logic for the y and z components:

𝑊=
1
2
2 2
𝑀⎡⎢𝑣𝑥2 + 𝑣𝑦2 + 𝑣𝑧2 − 𝑣𝑥1 + 𝑣𝑦1 + 𝑣𝑧1
⎣
2
( 2 2
)⎤⎥⎦
2

𝑊=
1
2 ( 2
𝑀 𝑣2 − 𝑣1 ; 𝑊 =
2
) 1
2
2
𝑚𝑣2 −
1
2
𝑚𝑣1
2

Thus,

1 2 1 2
𝑊= 2
𝑚𝑣2 − 2
𝑚𝑣1
𝑊 = 𝐾2 − 𝐾1 or ⇒𝑊 = Δ𝐾

→ →
This is known as the Work-Energy Theorem. For a particle with momentum 𝑝 = 𝑚𝑣 ,
1 2
the kinetic energy is: 𝐾= 2𝑚
𝑃

Motion in a Vertical Circle

Consider a particle of mass m attached to one end of a string, rotating in a vertical

circle of radius r with center O. As the particle moves from the lowest to the highest
point, its speed decreases due to gravitational acceleration, while in the reverse
direction, its speed increases.

If the particle is at point A, with the string making an angle θ with the vertical and
velocity v, the forces acting on it are:

1.​ Tension (T) in the string directed towards O

2.​ Weight (mg) acting downwards

The net force providing centripetal acceleration is:

 2
𝑚𝑣
𝑇 − 𝑚𝑔𝑐𝑜𝑠⁡θ = 𝑟

( )
2
𝑣
Thus, the tension in the string is: 𝑇 = 𝑚 𝑔𝑐𝑜𝑠⁡θ + 𝑟

If v0 represents the speed at the highest point, then as the particle descends through
a vertical height h, the velocity changes:

ℎ = 𝑟(1 + 𝑐𝑜𝑠⁡θ)
2 2 2
𝑣 = 𝑣0 + 2𝑔ℎ = 𝑣0 + 2𝑔𝑟(1 + 𝑐𝑜𝑠⁡θ)

At specific points in the circular motion:

◦
(i) At the highest point (θ = 180 ):

2 2
⎡𝑣 ⎤ ⎡𝑣 ⎤
𝑇𝑐 = 𝑚⎢ 𝑟0 + 𝑔𝑐𝑜𝑠⁡(180)⎥ = 𝑚⎢ 𝑟0 − 𝑔⎥
⎣ ⎦ ⎣ ⎦
For the string to remain taut, the tension must be zero or positive, implying that the
minimum velocity at the highest point is:

𝑣0 = 𝑟𝑔
◦
(ii) At the lowest point (θ = 0 ):

2
⎡𝑣 ⎤
𝑇𝐵 = 𝑚⎢ 𝑟𝐵 + 𝑔⎥
⎣ ⎦
Using energy conservation, the velocity at point B is:

2 2 2 2
(
𝑣𝐵 = 𝑣0 + 4𝑟𝑔 = 𝑟𝑔 + 4𝑟𝑔 = 5𝑟𝑔; 𝑢𝑠𝑖𝑛𝑔 𝑣 = 𝑢 + 2𝑔ℎ ; 𝑣𝐵 = 5𝑟𝑔 )
The minimum tension at point B is: TB = 6mg
 ◦
(iii) At point E (θ = 90 ):

2
𝑚𝑣𝐸
𝑇𝐸 = 𝑟

The velocity at point E is:

𝑉𝐸 = 3𝑟𝑔
Thus, the tension at point E is:

𝑇𝐸 = 𝑚( ) = 3𝑚𝑔
3𝑟𝑔
𝑟

In the case where the particle is not attached to a string but instead follows a circular
track of radius r, and has normal reaction N:

2
𝑚𝑣
𝑚𝑔𝑐𝑜𝑠⁡θ − 𝑁 = 𝑟

At the highest point:

2
𝑚𝑣
𝑚𝑔 − 𝑁 = 𝑟
When N = 0, the minimum speed is:

𝑣 = 𝑟𝑔

Condition for Completing the Loop ( 𝑢 ≥ 5𝑔𝑅)

For the particle to complete the circle, the string must remain taut at the highest
point (θ = π). For this to be true:

2
𝑚𝑣𝑚𝑖𝑛
𝑚𝑔 = 𝑅

2
or, 𝑣𝑚𝑖𝑛 = 𝑔𝑅

or, 𝑣𝑚𝑖𝑛 = 𝑔𝑅
From energy conservation:

𝑢𝑚𝑖𝑛 = 5𝑔𝑅

If 𝑢≥ 5𝑔𝑅 , the particle completes the loop. At this velocity, tension at the lowest

point reaches 6mg. If 𝑢 < 5𝑔𝑅 , the particle will not complete the circle, and
different outcomes are possible.

Condition of Leaving the Circle ( 2𝑔𝑅 < 𝑢 < 5𝑔𝑅)

If 𝑢 < 5𝑔𝑅,, then the tension in the string becomes zero before the particle reaches
the highest point. From Equation (iii), tension in the string becomes zero ( T = 0 )
when:

2 2
−𝑣 2𝑔ℎ−𝑢
𝑐𝑜𝑠⁡θ = 𝑅𝑔
or 𝑐𝑜𝑠⁡θ = 𝑅𝑔

Substituting this value of 𝑐𝑜𝑠 θ into Equation (i):

2 2
2𝑔ℎ−𝑢 ℎ 𝑢 +𝑅𝑔
𝑅𝑔
= 1 − 𝑅
⇒ ℎ = 3𝑔
= ℎ1 𝑠𝑎𝑦
Thus, at height h1, the tension in the string becomes
zero. Further, if 𝑢 < 5𝑔𝑅, the velocity of the particle
becomes zero when:

2
2 𝑢
0 = 𝑢 − 2𝑔ℎ or ℎ = 2𝑔
= ℎ2 (say)

This implies that the particle will leave the circle if tension alone in the string reaches
zero but the velocity does not (T = 0, but 𝑣≠ 0). This is possible only when:

2 2
𝑢 +𝑅𝑔 𝑢 2 2 2
ℎ1 < ℎ2 or 3𝑔
< 2𝑔
or 2𝑢 + 2𝑅𝑔 < 3𝑢 or 𝑢 > 2𝑅𝑔 or 𝑢 > 2𝑅𝑔.

Therefore, if 2𝑔𝑅 < 𝑢 < 5𝑔𝑅, the particle leaves the circular path.
From Equation (iv), we observe that h > R if u2 > 2Rg, which implies that the particle
◦ ◦
will leave the circle when h > R or, 90 < θ < 180 .

Condition of Oscillation (0 < 𝑢 ≤ 2𝑔𝑅)

The particle will oscillate only if the velocity reaches zero but the tension in the string
does not ( v = 0 , but 𝑇≠ 0 ). This is possible when:

h2 < h1

2 2
𝑢 𝑢 +𝑅𝑔
2𝑔
< 3𝑔

2 2
3𝑢 < 2𝑢 + 2𝑅𝑔

2
𝑢 < 2𝑅𝑔

𝑢 < 2𝑅𝑔

Moreover, if h1 = h2 at 𝑢 = 2𝑅𝑔, then both the tension and velocity become zero

simultaneously. From Equation (iv), we see that ℎ≤𝑅 if 𝑢≤ 2𝑅𝑔, meaning the
◦ ◦
particle oscillates within the lower half of the circle 0 < θ ≤ 90 .( )
Note: These conditions apply for a particle moving in a vertical circle attached to a
string. They also apply to a particle moving inside a smooth spherical shell of radius
R, with tension replaced by normal reaction (N).

For a particle of mass m attached to a light rod and moving in a vertical circle of
radius R, the minimum velocity at the bottommost point to complete the circle is not

5𝑔𝑅 . In this case, the velocity at the topmost point can be zero. By applying the
conservation of mechanical energy between points A and B.
 2 2
1
2 (
𝑚 𝑢 −𝑣 ) = 𝑚𝑔ℎ
1 2
or, 2
𝑚𝑢 = 𝑚𝑔(2𝑅)(𝑎𝑠𝑣 = 0)

∴ 𝑢 = 2 𝑔𝑅
Therefore, the minimum value of u in this case is 2 𝑔𝑅. The same applies when a
particle moves inside a smooth vertical tube as shown in below.

A Body Moving Inside a Hollow Tube

The discussion above applies to this scenario as well, but the tension in the string is
replaced by the normal reaction (N) from the surface. If N is the normal reaction at
the lowest point, then:

( )
2 2
𝑚𝑣1 𝑣1
𝑁 − 𝑚𝑔 = 𝑟
;𝑁 = 𝑚 𝑟
+ 𝑔
At the highest point of the circle:

( )
2 2
𝑚𝑣2 𝑣2
𝑁 + 𝑚𝑔 = 𝑟
𝑁= 𝑚 𝑟
− 𝑔

Thus, N ≥ 0 implies that:

𝑣1 ≥ 5 𝑔

Figure depicts a block moving inside a hollow sphere,

and all other equations can be derived by simply
replacing tension (T) with normal reaction (N).

Body Moving on a Spherical Surface

Consider a small body of mass m placed on the top of a smooth sphere of radius r. As
the body slides down the surface, we wish to determine the point at which it leaves
the surface.

At a certain point C, the forces acting on the mass

are the normal reaction (R) and the weight (mg).
The radial component of weight (mg cosθ) acts
toward the center. The centripetal force is given
by:

2
𝑚𝑣
𝑚𝑔𝑐𝑜𝑠⁡θ − 𝑅 = 𝑟
At the point where the body flies off the surface (R = 0):

2
𝑚𝑣
𝑔𝑐𝑜𝑠⁡ϕ − 𝑅 = 𝑟
Applying conservation of energy:

2
1 2 2 𝑣
2
𝑚𝑣 = 𝑚𝑔(𝐵𝑁) = 𝑚𝑔(𝑂𝐵 − 𝑂𝑁) = 𝑚𝑔𝑟(1 − 𝑐𝑜𝑠⁡ϕ)𝑣 = 2𝑟𝑔(1 − 𝑐𝑜𝑠⁡ϕ); 2(1 − 𝑐𝑜𝑠⁡ϕ) = 𝑟𝑔
From above Equations:
 𝑐𝑜𝑠⁡ϕ = 2 − 2𝑐𝑜𝑠⁡ϕ; 3𝑐𝑜𝑠⁡ϕ = 2
This denotes the angle at which the body leaves the surface. The height from the
ground at that point is:

(
= 𝐴𝑁 = 𝑟(1 + 𝑐𝑜𝑠⁡ϕ) = 𝑟 1 +
2
3 )= 5
3
𝑟

The Law of Conservation of Energy

In many physical processes, the sum of kinetic and potential energy is not conserved.
This is often due to the presence of dissipative forces such as friction.

A more comprehensive version of the law of conservation of energy was developed

by including other forms of energy, such as thermal, electrical, chemical, and nuclear
energies.

The change in all forms of energy can be described by the equation:

Δ𝐾𝐸 + Δ𝑈 + Δ (all other forms of energy) ≡0

This equation embodies the law of conservation of energy, which is one of the most
fundamental principles in physics. It states that:

"The total energy of a closed system remains constant—it can neither be created
nor destroyed, but it can be transformed from one form to another and
transferred between bodies."

Problem-Solving Tactics

(a) Identify Known and Unknowns: Start by isolating the known quantities and the
unknowns in the problem. Write down the relevant equations accordingly.

(b) Path from Unknown to Known: Determine a strategy to connect unknown

quantities to known ones using appropriate equations.
(c) Avoid Common Mistakes: Be mindful of common mistakes, such as errors in units
or incorrect conversion between units.

(d) Nature of Energy: Remember that energy is a scalar quantity. Always understand
clearly which entity is gaining or losing energy.

(e) Visualization: Physically visualize the problem, which will help build confidence in
solving the equations related to it.

(f) Use Integration Step-by-Step: For problems involving integration, solve the
problem step by step by considering each event independently.

(g) Master Special Cases: It is advisable to master special cases and boundary
conditions of circular motion, as many problems can be reduced to these special
cases after a few simplifications.