Introduction

PhysicsWork, Energy And PowerFor NEET aspirants

Work, energy and power are three linked scalar quantities that describe how forces transfer motion. Work is the dot product of force and displacement, $W = F \cdot s = F s cos θ$ , measured in joules. Energy is the capacity to do work, existing chiefly as kinetic energy $K = \frac{1}{2} m v^{2}$ and potential energy (gravitational $m g h$ or elastic $\frac{1}{2} k x^{2}$ ). Power is the rate of doing work, $P = d W / d t = F \cdot v$ , measured in watts. This concept builds the foundations for JEE Main, JEE Advanced and NEET, covering conservative forces, potential energy curves and the different forms of energy.

Key Formulas - Quick Reference

Work (constant force): $W = F \cdot s = F s cos θ$
Work (variable force): $W = \int_{A}^{B} F \cdot d s$ ; along $x$ : $W = \int_{x_{1}}^{x_{2}} F (x) d x$
Work graphically = area under the $F$ - $x$ curve
Kinetic energy: $K = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}$
Gravitational PE (near Earth): $U_{g} = m g h$
Elastic (spring) PE: $U_{s} = \frac{1}{2} k x^{2}$
Relation between force and PE: $F_{x} = - \frac{d U}{d x}$
Equilibrium condition: $\frac{d U}{d x} = 0$ ; stable if $\frac{d ^{2} U}{d x ^{2}} > 0$ , unstable if $< 0$ , neutral if $= 0$
Average power: $P_{av} = \frac{W}{t}$ ; instantaneous: $P = F \cdot v = F v cos θ$
Units: $1 J = 1 0^{7} erg$ , $1 cal = 4.2 J$ , $1 eV = 1.6 \times 1 0^{- 19} J$ , $1 HP = 746 W$

1. Work Done by a Force

Work is done by a force when its point of application moves. If a constant force $F$ produces a displacement $s$ making angle $θ$ with the force, then

W = F \cdot s = F s cos θ

Work is a scalar. SI unit is the joule (J), equal to one newton-metre. Only the component of force along the displacement contributes; the perpendicular component does no work.

Figure 1: Work by a force at angle

θ

. Only the component

F cos θ

along the displacement

s

does work; the perpendicular component

F sin θ

contributes nothing.

1.1 Positive, Negative and Zero Work

The sign of $cos θ$ decides the sign of the work.

Angle	Sign of work	Example
$0 \leq θ < 9 0^{\circ}$	Positive	Lifting force on a rising load; spring force on stretching hand
$θ = 9 0^{\circ}$	Zero	Tension in a pendulum string; normal reaction on a walking block; magnetic force on a moving charge
$9 0^{\circ} < θ \leq 18 0^{\circ}$	Negative	Gravity on a body being lifted; friction on a sliding block

Figure 2: Sign of work depends on the angle between force and displacement. Left: lifting force is along

s

, giving

W > 0

. Middle: pendulum tension is perpendicular to

s

, so

W = 0

. Right: gravity opposes upward

s

, so

W < 0

1.2 Work Done by Multiple Forces

If forces $F_{1}, F_{2}, F_{3}, \dots$ act simultaneously on a particle undergoing displacement $s$ , the net work equals the work done by the net force, which is also the algebraic sum of individual works:

W_{net} = (i \sum F_{i}) \cdot s = F_{1} \cdot s + F_{2} \cdot s + \dots = W_{1} + W_{2} + W_{3} + \dots

1.3 Important Properties of Work

Work is defined for an interval (or displacement). There is no concept of instantaneous work, unlike instantaneous velocity.
For a given displacement, the work done by a force is independent of the type of motion (uniform, accelerated, retarded).
If a body is in dynamic equilibrium (net force zero), the total work is zero, but individual forces may still do non-zero work.
Work done by a force in a given displacement is independent of other forces acting simultaneously.
A force is frame-independent, but displacement is not. Therefore work done by a force can differ in different reference frames.

Solved Example 1

A block of ice is drawn through 500 cm along a smooth horizontal surface by a rope pulling with 100 dyne at

3 0^{\circ}

above the horizontal. Find the work done.

Solution:

Only the horizontal component does work. $W = F s cos θ = 100 \times 500 \times cos 3 0^{\circ} = 100 \times 500 \times \frac{3}{2} \approx 43, 300 erg$

Solved Example 2 (Frame-dependence of work)

A 2 kg block sits on a flatcar accelerating at

1 m/s^{2}

. Over a

10 m

journey of the cart, find the work done by friction on the block (a) with respect to the driver, (b) with respect to a ground observer.

Solution:

(a) With respect to the driver, the block does not move, so displacement is zero and $W_{f} = 0$ .

(b) With respect to the ground, the block moves the same $10 m$ as the cart (it is at rest on the cart). Friction supplies the accelerating force: $f = ma = 2 \times 1 = 2 N$ , directed along motion. $W_{f} = f \cdot s = 2 \times 10 = 20 J$ Same force, different frame, different work.

2. Work Done by a Variable Force

When the force changes with position, we split the path into infinitesimal displacements $d s$ over which the force is essentially constant, then integrate:

d W = F \cdot d s, W_{A \to B} = \int_{A}^{B} F \cdot d s

If $F = F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k}$ and $d s = d x \hat{i} + d y \hat{j} + d z \hat{k}$ , then

W = \int_{x_{A}}^{x_{B}} F_{x} d x + \int_{y_{A}}^{y_{B}} F_{y} d y + \int_{z_{A}}^{z_{B}} F_{z} d z

2.1 Graphical Method: Area under the F-x Curve

For one-dimensional motion, the work done by a variable force $F (x)$ from $x_{1}$ to $x_{2}$ equals the area under the $F$ versus $x$ graph between those limits. Area above the axis counts as positive work; area below counts as negative.

Figure 2: Work as area under the

F

x

curve. Rectangle (0 to 4 m) plus triangle (4 to 6 m) gives total work

W = 20 + 5 = 25 J

Solved Example 3

A force

F = 0.5 x + 10

(newton, with

x

in metre) acts on a particle. Find the work done as the particle moves from

x = 0

x = 2 m

Solution:

$W = \int_{0}^{2} F d x = \int_{0}^{2} (0.5 x + 10) d x = [\frac{0.5 x ^{2}}{2} + 10 x]_{0}^{2} = 1 + 20 = 21 J$

Solved Example 4 (Force in the xy plane)

A particle in the

x y

plane undergoes displacement

s = (2 \hat{i} + 3 \hat{j}) m

under a constant force

F = (5 \hat{i} + 2 \hat{j}) N

. Find the work done.

Solution:

$W = F \cdot s = (5) (2) + (2) (3) = 10 + 6 = 16 J$

2.2 Force as a Function of Time

If the force is given as a function of time, first find $v (t)$ from $F = m d v / d t$ , then use $d W = F v d t$ .

Solved Example 5

A force

F = 2 t^{2}

(newton) acts on a 2 kg block initially at rest. Find the work done by this force in 2 seconds.

Solution:

From $F = ma$ : $2 t^{2} = 2 a \Rightarrow a = t^{2}$ . Integrating with $v (0) = 0$ : $v = t^{3} /3$ . $d W = F d x = F v d t = 2 t^{2} \cdot \frac{t ^{3}}{3} d t = \frac{2}{3} t^{5} d t$ $W = \frac{2}{3} \int_{0}^{2} t^{5} d t = \frac{2}{3} \cdot \frac{2 ^{6}}{6} = \frac{64}{9} J \approx 7.11 J$

2.3 Internal Work

When a person pushes off a wall to move backward, the wall exerts a horizontal force but does no external work (the wall does not move). Yet the person accelerates and gains kinetic energy. The resolution: the person is a composite system, and internal forces between muscles, bones and joints do work by moving relative to each other. This is called internal work, and it can change the kinetic energy of a system even when no external work is done.

Rule of thumb. The work done by any force is calculated using the displacement of the point of application of that force in the chosen reference frame, not the displacement of the body's centre of mass.

3. Units and Dimensions of Work, Energy and Power

Work and energy share the same units and dimensions, because energy is measured by the work it can do. Power has an extra factor of $1/ time$ .

Quantity	Dimension	SI unit	CGS unit	Practical units
Work / Energy	$[M L^{2} T^{- 2}]$	joule (J)	erg	eV, kWh, calorie
Power	$[M L^{2} T^{- 3}]$	watt (W)	erg/s	HP, kW, MW

3.1 Useful Conversions

$1 J = 1 0^{7} erg$
$1 cal = 4.2 J$
$1 kWh = 3.6 \times 1 0^{6} J$
$1 eV = 1.6 \times 1 0^{- 19} J$
$1 HP = 746 W$
$1 MW = 1 0^{6} W$

4. Energy: Kinetic and Potential

Energy is the capacity to do work. Work and energy are mutually convertible: when a body does work, it loses energy; when work is done on a body, it gains energy. Mechanical energy has two forms - kinetic (from motion) and potential (from position or configuration).

4.1 Kinetic Energy

Kinetic energy is the energy possessed by a body by virtue of its motion. For a body of mass $m$ moving with speed $v$ :

K = \frac{1}{2} m v^{2}

Since $m$ and $v^{2}$ are always positive, kinetic energy is always non-negative and does not depend on the direction of motion. In terms of linear momentum $p = m v$ :

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

Solved Example 6

Two bodies of masses

m

and

4 m

have equal kinetic energies. Compare their linear momenta.

Solution:

Using $K = p^{2} / (2 m)$ , equal $K$ gives $p_{1}^{2} / (2 m) = p_{2}^{2} / (2 \cdot 4 m)$ , so $p_{2}^{2} = 4 p_{1}^{2}$ , i.e. $p_{2} = 2 p_{1}$ . The heavier body carries twice the momentum for the same kinetic energy.

4.2 Potential Energy

Potential energy is the energy stored by virtue of position, configuration or state of strain. The change in potential energy is related to work done by the (conservative) force through

W_{conservative} = - Δ U

Potential energy is a function of position only (not velocity or time).
Its value at a point depends on a chosen reference level; only differences $Δ U$ have physical meaning.
The difference $U_{B} - U_{A}$ is frame-independent (unlike absolute values that depend on choice of reference).
Potential energy can be defined only for conservative forces - for non-conservative forces the work depends on path, so no unique $U$ can be assigned.

Gravitational potential energy (near Earth)

For a mass $m$ at height $h$ above a chosen reference level (with $h ≪ R_{Earth}$ ):

U_{g} = m g h

Below the reference, $U_{g}$ is negative.

Elastic (spring) potential energy

For a spring of natural length $ℓ_{0}$ and force constant $k$ , stretched or compressed by $x$ from its natural length, the restoring force is $F_{s} = - k x$ (Hooke's law). Work done by the spring on a block that moves from $0$ to $x_{0}$ :

W_{spring} = \int_{0}^{x_{0}} (- k x) d x = - \frac{1}{2} k x_{0}^{2}

Therefore the energy stored (potential energy) equals the positive of the work done against the spring:

U_{s} = \frac{1}{2} k x^{2}

This is always non-negative, whether the spring is stretched or compressed (since $x^{2} \geq 0$ ). Zero of $U_{s}$ is at the natural length.

Figure 5: Elastic potential energy

U_{s} = \frac{1}{2} k x^{2}

is a symmetric parabola. Both stretching (

x > 0

) and compression (

x < 0

) give positive

U_{s}

, with the minimum (

U_{s} = 0

) at the natural length.

Solved Example 7 (Spring on rough surface)

A spring of force constant

k

is attached to a wall and its free end pushes a block of mass

m

resting on a rough horizontal surface with friction coefficient

μ

. Find the maximum energy that can be stored in the spring for which the block remains stationary.

Solution:

The block is on the verge of sliding when the spring force equals limiting friction: $k x_{m a x} = μ m g \Rightarrow x_{m a x} = \frac{μ m g}{k}$ Maximum stored elastic PE: $U_{m a x} = \frac{1}{2} k x_{m a x}^{2} = \frac{1}{2} k (\frac{μ m g}{k})^{2} = \frac{μ ^{2} m ^{2} g ^{2}}{2 k}$

Solved Example 8 (Two-particle PE, finding minimum)

The potential energy of a two-particle system is

U (x) = \frac{a}{x ^{2}} - \frac{b}{x}

, where

x

is the separation and

a, b > 0

. Find the minimum value of

U

Solution:

For a minimum, $d U / d x = 0$ : $\frac{d U}{d x} = - \frac{2 a}{x ^{3}} + \frac{b}{x ^{2}} = 0 \Rightarrow x = \frac{2 a}{b}$ Substituting back: $U_{m i n} = \frac{a}{( 2 a / b ) ^{2}} - \frac{b}{( 2 a / b )} = \frac{a b ^{2}}{4 a ^{2}} - \frac{b ^{2}}{2 a} = - \frac{b ^{2}}{4 a}$ The minimum PE is $- b^{2} / (4 a)$ .

5. Conservative and Non-Conservative Forces

A force is conservative if the work it does on a particle moving between two points depends only on the endpoints, not on the path taken. Equivalently, its work over any closed loop is zero.

Figure 3: Work done against gravity in raising a mass to height

h

m g h

along every path (a), (b) or (c). Gravity is a conservative force - only the change in height matters.

5.1 Examples of Conservative Forces

Gravitational force (both near-Earth and Newton's universal form)
Elastic (spring) force
Electrostatic force between charges
All central forces (force along the line joining two centres, magnitude depending only on separation)

5.2 Non-Conservative Forces

A force is non-conservative if the work it does depends on the path taken between the endpoints. The work over a closed loop is not zero. Common examples:

Kinetic friction - work equals $- μ_{k} N \cdot (path length)$ , which grows with the length of the path travelled.
Air resistance and other viscous forces (velocity-dependent).
Applied external forces such as a push or pull (they generally cannot be derived from a potential).

Solved Example 9 (Path-dependence)

Calculate the work done by the force

F = y \hat{i}

in moving a particle from

(0, 0)

(1, 1)

along (a)

y = x

, (b)

y = x^{2}

Solution:

Along the given path, $d W = F \cdot d s = y d x$ .

(a) On $y = x$ : $W = \int_{0}^{1} x d x = \frac{1}{2} J$ .

(b) On $y = x^{2}$ : $W = \int_{0}^{1} x^{2} d x = \frac{1}{3} J$ .

The two answers differ, so $F = y \hat{i}$ is non-conservative.

Figure 7: Two paths from

(0, 0)

(1, 1)

give different work under the force

F = y \hat{i}

\frac{1}{2} J

along

y = x

and

\frac{1}{3} J

along

y = x^{2}

. Path-dependence marks the force as non-conservative.

5.3 Conservative vs Non-Conservative: Summary

Property	Conservative force	Non-conservative force
Work over a path	Depends only on endpoints	Depends on the path
Work in a closed loop	Zero	Non-zero
Potential energy $U$	Well defined; $F = - d U / d x$	Cannot be defined
Mechanical energy	Conserved (when only these forces act)	Not conserved; energy dissipates as heat, sound, etc.
Recoverability of work	Fully recoverable	Not fully recoverable
Examples	Gravity, spring, electrostatic	Friction, viscous drag, applied push

6. Force from Potential Energy and PE Curves

For a conservative force in one dimension, the force is the negative slope of the potential energy:

F (x) = - \frac{d U}{d x}

In three dimensions, the force is the negative gradient of $U$ :

F = - \nabla U = - (\frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} + \frac{\partial U}{\partial z} \hat{k})

Wherever $U (x)$ has a slope, a conservative force pushes the particle "downhill" on the PE curve.

6.1 Equilibrium and its Types

A particle is in equilibrium where the net force is zero, i.e. $d U / d x = 0$ . Whether that equilibrium is stable depends on the curvature of $U$ .

Figure 4: Potential energy curve

U (x)

. At P and R (troughs)

d^{2} U / d x^{2} > 0

- stable equilibrium. At Q (peak)

d^{2} U / d x^{2} < 0

- unstable equilibrium. On the flat plateau S,

d^{2} U / d x^{2} = 0

- neutral equilibrium.

Type	Conditions	Behaviour after small displacement
Stable	$\frac{d U}{d x} = 0, \frac{d ^{2} U}{d x ^{2}} > 0$ (U is a minimum)	Restoring force brings particle back
Unstable	$\frac{d U}{d x} = 0, \frac{d ^{2} U}{d x ^{2}} < 0$ (U is a maximum)	Force pushes particle further away
Neutral	$\frac{d U}{d x} = 0, \frac{d ^{2} U}{d x ^{2}} = 0$ (U is constant)	Particle stays in the new position

Solved Example 10 (Lennard-Jones-type PE)

The potential energy between two atoms in a molecule is

U (x) = \frac{a}{x ^{12}} - \frac{b}{x ^{6}}

, where

a, b > 0

. Find the position of stable equilibrium.

Solution:

For equilibrium, $d U / d x = 0$ : $\frac{d U}{d x} = - \frac{12 a}{x ^{13}} + \frac{6 b}{x ^{7}} = 0 \Rightarrow x^{6} = \frac{12 a}{6 b} = \frac{2 a}{b} \Rightarrow x = (\frac{2 a}{b})^{1/6}$ At this $x$ , one can verify $d^{2} U / d x^{2} > 0$ , so it is a stable equilibrium (the bond length of the molecule).

Solved Example 11 (Parabolic PE)

The PE of a conservative system is

U (x) = a x^{2} - b x

, with

a, b > 0

. Find the equilibrium position and its type.

Solution:

$F = - d U / d x = - (2 a x - b) = b - 2 a x$ . Setting $F = 0$ : $x = b / (2 a)$ . Also $d^{2} U / d x^{2} = 2 a > 0$ , so $U$ is minimum here. Therefore $x = b / (2 a)$ is a position of stable equilibrium.

7. Power

Power is the rate at which work is done by (or energy is transferred by) a force.

7.1 Average and Instantaneous Power

P_{av} = \frac{W}{t} = \frac{total work done}{total time taken}

P_{inst} = \frac{d W}{d t} = F \cdot v = F v cos θ

where $θ$ is the angle between force and velocity. Power is a scalar.

7.2 Units and Dimensions

SI unit is watt (W): $1 W = 1 J/s = 1 N \cdot m/s$ . Dimensional formula: $[M L^{2} T^{- 3}]$ .

$1 kW = 1 0^{3} W$
$1 MW = 1 0^{6} W$
$1 HP = 746 W$ (standard definition, $1 HP = 550 ft-lb/s$ )

Solved Example 12 (Engine pulling a train)

An engine of mass 20 tons pulls 20 wagons, each of mass 20 tons, at a constant

72 km/hr

on level track. Coefficient of kinetic friction is

μ_{k} = 0.01

. Find the power developed by the engine. (Take

g = 10 m/s^{2}

Solution:

Total mass (engine + wagons) $= 21 \times 20 = 420$ tons $= 420 \times 1 0^{3} kg$ . At constant velocity, engine force equals friction: $F = μ_{k} m g$ . Speed $v = 72 km/hr = 20 m/s$ . $P = F v = μ_{k} m g v = 0.01 \times (420 \times 1 0^{3}) \times 10 \times 20 = 8.4 \times 1 0^{5} W = 840 kW$

Solved Example 13 (Pumping water)

A 1 kW motor pumps water from a 10 m deep well. Calculate the mass of water pumped per second. (Take

g = 9.8 m/s^{2}

Solution:

In 1 second, work done = $1 0^{3} J$ . For a mass $m$ lifted through 10 m: $W = m g h$ . So $m = \frac{P t}{g h} = \frac{1 0 ^{3} \times 1}{9.8 \times 10} \approx 10.2 kg/s$

Solved Example 14 (Comparing two workers)

A coolie takes 60 s to lift a box through 2 m. Another does the same job in 30 s. Compare their powers and total energies expended.

Solution:

Same work $W = m g h$ done by both, so total energy expended is equal. But $\frac{P _{2}}{P _{1}} = \frac{t _{1}}{t _{2}} = \frac{60}{30} = 2$ The second coolie has twice the power of the first.

8. Different Forms of Energy

Beyond mechanical energy, physics recognises several other forms that can interconvert. The law of conservation of energy states that the total energy of an isolated system, summed over all forms, is constant.

Form	Nature / typical formula	Example
Kinetic	$K = \frac{1}{2} m v^{2}$	A moving car, a flowing river
Potential (gravitational, elastic, electric)	$m g h, \frac{1}{2} k x^{2}, q V$	Stretched spring, water in a dam
Thermal (heat)	Random KE of molecules; $Q = m c Δ T$	Hot object, sliding friction generating heat
Chemical	Energy stored in molecular bonds	Food, fuel, batteries
Electrical	Energy of moving charges; $\frac{1}{2} C V^{2}$ , $\frac{1}{2} L I^{2}$	Household current, lightning
Radiant / electromagnetic	Energy of photons; $E = h ν$	Sunlight, radio waves
Nuclear	Binding energy of nucleus	Fission (reactors), fusion (Sun)
Sound	Mechanical wave in a medium	Musical instruments

8.1 Mass-Energy Equivalence

Einstein's special theory of relativity showed that mass is itself a form of energy:

E = m c^{2}

Here $c = 3 \times 1 0^{8} m/s$ is the speed of light. A tiny amount of mass corresponds to an enormous amount of energy - this is why nuclear reactions (fission, fusion) release so much energy for small mass differences.

Numerical feel.

1 u

(atomic mass unit)

= 1.66 \times 1 0^{- 27} kg

corresponds to about

931 MeV

of energy. This unit is central to nuclear physics chapters.

Common Mistakes to Avoid

Watch out

Confusing $cos θ$ signs. For work, $θ$ is the angle between force and displacement, not between force and some arbitrary axis. If they point in opposite directions ( $θ = 18 0^{\circ}$ ), $cos θ = - 1$ , and work is negative.
Assuming zero net force means zero work by each force. A book pushed horizontally at constant velocity - applied force does positive work, friction does equal negative work; net work is zero, but individual works are not.
Forgetting frame-dependence. Work done by friction on a block resting on an accelerating cart is zero for the driver but non-zero for a ground observer.
Using $W = F s$ blindly. Only if $F$ is constant and along $s$ . For variable force, integrate; for angled force, use $F s cos θ$ .
Elastic PE with negative $x$ . $U_{s} = \frac{1}{2} k x^{2}$ is always positive - it does not matter whether the spring is stretched ( $x > 0$ ) or compressed ( $x < 0$ ).
Reference level for gravitational PE. The value of $U_{g}$ depends on where you place the zero. Only differences in $U_{g}$ matter for physics.
Second-derivative sign for equilibrium type. At $d U / d x = 0$ : $d^{2} U / d x^{2} > 0$ means stable (U minimum), not unstable. Students often flip this.
Trying to define PE for friction. Non-conservative forces do path-dependent work; no potential energy function exists.
Confusing $1 HP = 746 W$ with metric HP. The Indian engineering syllabus uses $1 HP = 746 W$ ; the metric "PS" (735.5 W) is different and not needed for JEE/NEET.
Power = work / time only for constant power. For time-varying power, use $P = d W / d t$ , and integrate to get total work: $W = \int P d t$ .

Frequently Asked Questions

Q1. What is the difference between work and energy in physics?

Work is the transfer of energy by a force acting through a displacement; it is a process. Energy is the stored capacity to do work; it is a property of a system at a given instant. Both are scalars measured in joules, and work done on a body raises its energy while work done by a body lowers it.

Q2. When is work done by a force equal to zero?

Work is zero in three situations: (1) when the displacement is zero (a person holding a suitcase but not walking), (2) when the force is zero, and (3) when force is perpendicular to displacement (tension in a pendulum string, magnetic force on a moving charge, normal reaction during horizontal motion).

Q3. Can kinetic energy be negative?

No. Kinetic energy $K = \frac{1}{2} m v^{2}$ contains $m > 0$ and $v^{2} \geq 0$ , so $K \geq 0$ always. It does not depend on the direction of motion. Change in kinetic energy $Δ K$ can of course be negative (when the body slows down).

Q4. Can potential energy be negative?

Yes. Potential energy depends on the chosen reference level. For gravitational PE $U_{g} = m g h$ , if the body is below the reference level, $h$ and hence $U_{g}$ are negative. Only differences $Δ U$ are physically meaningful; the absolute value has no independent meaning.

Q5. Why can potential energy be defined only for conservative forces?

PE is defined so that the work done by the force equals $- Δ U$ . For this to give a unique value at every point, the work must depend only on endpoints, not on the path taken. Non-conservative forces like friction give different work on different paths, so no single-valued PE function can be assigned.

Q6. What is the relation between force and potential energy?

For a conservative force in one dimension, $F (x) = - \frac{d U}{d x}$ . The force points along the direction in which $U$ decreases - the particle is pushed 'downhill' on the PE curve. In 3D, $F = - \nabla U$ .

Q7. How do you identify stable, unstable and neutral equilibrium from a PE graph?

All equilibrium positions satisfy $d U / d x = 0$ (flat tangent). At a minimum ( $d^{2} U / d x^{2} > 0$ ) the equilibrium is stable - like the bottom of a valley. At a maximum ( $d^{2} U / d x^{2} < 0$ ) it is unstable - like a ball balanced on a hilltop. On a flat plateau ( $d^{2} U / d x^{2} = 0$ ) it is neutral.

Q8. What is the difference between average and instantaneous power?

Average power is total work divided by total time, $P_{av} = W / t$ . Instantaneous power is the rate at a specific moment, $P = d W / d t = F \cdot v$ . For constant power the two are equal; for varying power they differ, and instantaneous power is what you measure at any given instant.

Q9. Why is 1 HP equal to 746 W?

1 horsepower was historically defined as 550 foot-pounds per second (the estimated rate at which a strong horse could work). Converting to SI: $550 \times 0.3048 m \times 4.448 N/s \approx 746 W$ . This is the definition used in JEE and NEET problems.

Q10. What is mass-energy equivalence and where does it appear in JEE/NEET?

Einstein's relation $E = m c^{2}$ states that mass is a form of energy. In JEE and NEET it appears in Modern Physics (nuclear reactions, binding energy, radioactive decay, defect mass in fusion and fission). $1 u = 1.66 \times 1 0^{- 27} kg$ corresponds to about $931 MeV$ .

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