Work, Energy and Power Notes for NEET — Class 11 NCERT

Q: How many questions come from Work, Energy and Power in NEET 2027?

You can expect 3 to 5 questions from Work, Energy and Power in NEET 2027. The chapter has the highest PYQ frequency in Class 11 Physics. Work-energy theorem, spring energy, conservation of energy, power and 1D collisions are tested almost every year.

Q: Why is Work, Energy and Power so important for NEET?

Energy methods often solve problems that look impossible by force methods. Once you have the work-energy theorem, you can answer questions about the speed of a body at a given point without ever computing the force on it. NEET examiners love problems that mix springs, inclined planes, friction and circular motion — all easier with energy.

Q: What is the work-energy theorem?

The net work done on a body equals its change in kinetic energy. In symbols, W_net = (1/2)mv squared minus (1/2)mu squared. This single statement replaces a long sequence of force and acceleration calculations.

Q: What is the formula for the energy stored in a spring?

A spring stretched (or compressed) from its natural length by x stores potential energy U = (1/2)kx squared, where k is the spring constant. Note the square: doubling the extension quadruples the stored energy.

Q: What is the difference between average power and instantaneous power?

Average power equals total work divided by total time, P_avg = W over t. Instantaneous power is the rate at that instant, P = F dot v. They are equal when the force and speed are constant; they differ when either changes.

Q: In a perfectly inelastic 1D collision, how much kinetic energy is lost?

For one body of mass m1 with velocity u colliding with another mass m2 at rest, the kinetic energy lost is (m1 m2 over m1 plus m2) times (u squared over 2). Maximum loss happens when the masses are equal — half the original KE is converted to heat or deformation.

Q: What is the minimum speed at the top of a vertical circular loop?

For a body moving on the inside of a vertical loop of radius r, the minimum speed at the top is the square root of (g r). At this speed, gravity alone provides the centripetal force; the string or track tension is zero. Below this speed, the body falls off the loop.

Introduction

Work, Energy and Power is the chapter where physics gets economical. Many problems that look hopeless using Newton's laws collapse into a single energy equation. If you can compute kinetic energy at the start, add the work done by every force in between and read off the kinetic energy at the end, you have done the entire problem.

For NEET 2027 this is the most heavily tested chapter in Class 11 Physics. Expect 3 to 5 questions per year on the work-energy theorem, spring energy, conservation of energy, power and 1D collisions. Vertical circular motion and inclined-plane-with-friction problems also lean on this chapter.

Work done by a constant force

When a constant force $F$ acts on a body that undergoes a displacement $d$ , the work done is the dot product:

W = F \cdot d = F d cos θ,

where $θ$ is the angle between $F$ and $d$ . Work is a scalar with SI unit joule (J), where $1 J = 1 N \cdot m$ . Dimensions: $[M L^{2} T^{- 2}]$ .

Sign of work

$θ < 90°$ → $W > 0$ : force has a component along motion.
$θ = 90°$ → $W = 0$ : force perpendicular to motion does no work. Centripetal force is the classic example.
$θ > 90°$ → $W < 0$ : force opposes motion. Friction usually does negative work.

Force F: 20 N

Displacement d: 5 m

Angle θ between F and d: 60°

Work done

W = 50.00 J

W = F d cos θ = 20 \times 5 \times cos 60° = 50.00 J

Force has a component along motion → positive work, body gains KE.

Work done by a variable force

When the force changes with position, the work is the integral of $F d x$ along the path:

W = \int_{x_{1}}^{x_{2}} F (x) d x .

Geometrically, the work equals the area under the F-x graph between the two positions. This is the most-tested form of the variable-force idea on NEET.

Slope/value k: 4

Final position x: 5 m

Work = signed area

W = 50.00 J

F = k·x (driving)

Kinetic energy

The kinetic energy of a particle of mass $m$ moving with speed $v$ is:

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

Kinetic energy is always non-negative and depends on the frame of reference. It can also be written in terms of momentum: $K = p^{2} / (2 m)$ . This form is extremely useful in collision problems.

The work-energy theorem

The single most useful tool in this chapter:

W_{net} = Δ K = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} .

The net work done by all forces on a body equals its change in kinetic energy. This holds regardless of the path or the time taken, as long as you account for every force. It is true even when the forces vary or the path is curved.

When to reach for it

The question asks for the speed at a particular point.
You know the work done by each force but the motion is complicated.
You need the work done by friction over an unknown path of known length.

Mass m: 2 kg

Initial speed u: 0 m/s

Final speed v: 10 m/s

Initial KE

0.00 J

Final KE

100.00 J

Net work needed

W_net = 100.00 J

Positive — net force in the direction of motion.

W_{net} = Δ K = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} = 100.00 - 0.00 = 100.00 J

Potential energy

Potential energy is energy stored in the configuration of a system. It only exists for conservative forces— forces whose work depends only on the start and end points, not the path. The two main NEET examples are gravity (near Earth's surface) and a spring.

Gravitational potential energy (near Earth)

For a body of mass $m$ at height $h$ above a chosen reference level:

U_{g} = m g h .

The reference level is your choice. Only differences in $U$ have physical meaning, so the answer to a problem never depends on where you put the zero.

Conservative force ↔ potential energy

For a 1D conservative force, $F = - \frac{d U}{d x}$ . The negative sign says the force pushes the body toward lower potential energy.

Practice these on the timed test

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Elastic (spring) potential energy

Hooke's law for an ideal spring: a spring stretched (or compressed) from its natural length by $x$ exerts a restoring force:

F = - k x,

where $k$ is the spring constant (units N/m). The work done on the spring to deform it from natural length to extension $x$ is the area of a triangle on the F-x graph:

U_{spring} = \frac{1}{2} k x^{2} .

Note the square: doubling the extension quadruples the stored energy.

Spring constant k: 200 N/m

Extension x: 0.10 m (10 cm)

Stored energy

1.000 J

Restoring force

20.00 N

U = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.10)^{2} = 1.000 J

F = k x = 200 \times 0.10 = 20.00 N

Conservation of mechanical energy

If only conservative forces do work on a system, its total mechanical energy is conserved:

K_{1} + U_{1} = K_{2} + U_{2} .

For a body in free fall from height $h$ , the speed at the bottom is $v = 2 g h$ , because all of $m g h$ converts into $\frac{1}{2} m v^{2}$ .

When energy is not conserved

If non-conservative forces (friction, air drag, deformation) do work, mechanical energy is lost as heat or sound:

Δ (K + U) = W_{non-conservative} .

For example, a block sliding down a rough incline ends with less $K$ than $m g h$ by an amount equal to the work done against friction.

Drop a ball of mass m from height H. As it falls, gravitational PE converts to KE while the total stays constant. Slide the height to see the energy mix at any instant.

Initial height H: 20 m

Mass m: 1 kg

Current height h: 20.0 m (above ground)

Speed at this height: 0.00 m/s

Energy budget (total = 200.00 J)

PE 100%

Kinetic

0.00 J

Potential

200.00 J

K E + U = const = m g H = 200.00 J

Power

Power is the rate at which work is done. Two flavours appear in NEET:

P_{avg} = \frac{W}{t}, P_{inst} = \frac{d W}{d t} = F \cdot v .

SI unit: watt (W), where $1 W = 1 J/s$ . Practical conversions you should know: $1 kW = 1 0^{3} W$ , $1 HP \approx 746 W$ , and $1 kWh = 3.6 \times 1 0^{6} J$ .

Instantaneous: car at constant speed

Driving force F: 500 N

Speed v: 20 m/s

P = F·v

10000 W = 10.00 kW = 13.40 HP

Average: pump lifting water

Mass m: 300 kg

Height h: 20 m

Time t: 60 s

P_avg = mgh / t

1000 W = 1.00 kW

P_{inst} = F v = 500 \times 20 = 10000 W

P_{avg} = \frac{m g h}{t} = \frac{300 \times 10 \times 20}{60} = 1000 W

Collisions in one dimension

Conservation of momentum always holds in collisions (no external impulsive forces). Whether kinetic energy is conserved tells you what kind of collision it is.

Perfectly elastic 1D collision

Two bodies, masses $m_{1}$ and $m_{2}$ , with initial velocities $u_{1}$ and $u_{2}$ . Momentum and kinetic energy both conserved. Solving the two equations gives:

v_{1} = \frac{( m _{1} - m _{2} ) u _{1} + 2 m _{2} u _{2}}{m _{1} + m _{2}}, v_{2} = \frac{( m _{2} - m _{1} ) u _{2} + 2 m _{1} u _{1}}{m _{1} + m _{2}} .

Special case: equal masses with one at rest → the velocities are exchanged. NEET tests this every other year.

Perfectly inelastic 1D collision

The two bodies stick together. Final common velocity:

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}} .

Kinetic energy lost (when the second mass is at rest, $u_{2} = 0$ ):

Δ K = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} u_{1}^{2} .

With equal masses, exactly half of the original kinetic energy is lost.

Vertical circular motion

A mass attached to a string (or a bead on the inside of a vertical loop) traces a circle in a vertical plane. Two questions matter for NEET:

Speed at the top vs at the bottom

Apply conservation of energy between the bottom (height 0) and the top (height $2 r$ ):

\frac{1}{2} m v_{bot}^{2} = \frac{1}{2} m v_{top}^{2} + m g (2 r) \Rightarrow v_{bot}^{2} = v_{top}^{2} + 4 g r .

Minimum speed at the top

At the top of the loop, both gravity and tension point toward the centre. The minimum speed occurs when the tension is zero and gravity alone provides the centripetal force:

m g = \frac{m v _{top,min}^{2}}{r} \Rightarrow v_{top,min} = g r .

Substituting into the energy equation gives the minimum speed at the bottom: $v_{bot,min} = 5 g r$ .

Loop radius r: 2 m

Speed at bottom v_bot: 10 m/s

Min v at top

4.47 m/s

Min v at bottom

10.00 m/s

Loop status

Completes the loop. Speed at top = 4.47 m/s

v_{top,min} = g r = 10 \times 2 = 4.47 m/s

v_{bot,min} = 5 g r = 100 = 10.00 m/s

v_{top}^{2} = v_{bot}^{2} - 4 g r = 100 - 80 = 20.00

Worked NEET problems

NEET-style problem · Work-energy theorem

Question

A body of mass

2 kg

initially at rest is acted on by a constant force of

10 N

over

5 m

. Take the surface as frictionless. Find the final speed.

Solution

By the work-energy theorem, $W_{net} = \frac{1}{2} m v^{2}$ :

(10) (5) = \frac{1}{2} (2) v^{2} \Rightarrow v^{2} = 50 \Rightarrow v = 52 m/s .

NEET-style problem · Spring energy

Question

A spring of force constant

k = 200 N/m

is compressed by

x = 0.10 m

. The energy stored in the spring is:

Solution

U = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.10)^{2} = \frac{1}{2} (200) (0.01) = 1 J .

NEET-style problem · Conservation of energy

Question

A ball of mass

0.5 kg

is dropped from a height of

20 m

. Take

g = 10 m/s^{2}

. Ignoring air resistance, find its speed just before impact.

Solution

Conservation of energy: $m g h = \frac{1}{2} m v^{2}$ :

v = 2 g h = 2 \times 10 \times 20 = 400 = 20 m/s .

Mass cancels — anything dropped from 20 m hits the ground at 20 m/s (ignoring drag).

NEET-style problem · Power

Question

A car of mass

1000 kg

moves at constant velocity

20 m/s

on a level road. The total resistive force is

500 N

. Find the power delivered by the engine.

Solution

At constant velocity, the driving force equals the resistive force: $F = 500 N$ .

P = F v = 500 \times 20 = 10000 W = 10 kW .

NEET-style problem · Vertical loop

Question

A small body of mass

m

is attached to a string of length

0.5 m

and whirled in a vertical circle. Take

g = 10 m/s^{2}

. The minimum speed at the top of the circle is:

Solution

v_{top,min} = g r = 10 \times 0.5 = 5 m/s \approx 2.24 m/s .

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Summary cheat sheet

Work (constant force): $W = F d cos θ$ .
Work (variable force): $W = \int F d x$ = area under F-x graph.
Kinetic energy: $K = \frac{1}{2} m v^{2} = p^{2} / (2 m)$ .
Work-energy theorem: $W_{net} = Δ K$ .
Gravitational PE: $U_{g} = m g h$ .
Spring PE: $U = \frac{1}{2} k x^{2}$ .
Conservation: $K_{1} + U_{1} = K_{2} + U_{2}$ (no non-conservative work).
Power: $P_{avg} = W / t$ , $P_{inst} = F \cdot v$ .
Elastic 1D collision (equal masses, one at rest): velocities exchange.
Inelastic 1D collision (KE lost, target at rest): $Δ K = \frac{m _{1} m _{2}}{2 ( m _{1} + m _{2} )} u_{1}^{2}$ .
Vertical loop minimum speeds: top $g r$ , bottom $5 g r$ .

Next: try the interactive widgets for work, spring energy, the work-energy theorem and the vertical loop, or work through the 30+ NEET PYQs with full solutions. To time yourself, take the free 10-question mock test.

Frequently asked questions

How many questions come from Work, Energy and Power in NEET 2027?

You can expect 3 to 5 questions from Work, Energy and Power in NEET 2027. The chapter has the highest PYQ frequency in Class 11 Physics. Work-energy theorem, spring energy, conservation of energy, power and 1D collisions are tested almost every year.

Why is Work, Energy and Power so important for NEET?

Energy methods often solve problems that look impossible by force methods. Once you have the work-energy theorem, you can answer questions about the speed of a body at a given point without ever computing the force on it. NEET examiners love problems that mix springs, inclined planes, friction and circular motion — all easier with energy.

What is the work-energy theorem?

The net work done on a body equals its change in kinetic energy. In symbols, W_net = (1/2)mv squared minus (1/2)mu squared. This single statement replaces a long sequence of force and acceleration calculations.

What is the formula for the energy stored in a spring?

A spring stretched (or compressed) from its natural length by x stores potential energy U = (1/2)kx squared, where k is the spring constant. Note the square: doubling the extension quadruples the stored energy.

What is the difference between average power and instantaneous power?

Average power equals total work divided by total time, P_avg = W over t. Instantaneous power is the rate at that instant, P = F dot v. They are equal when the force and speed are constant; they differ when either changes.

In a perfectly inelastic 1D collision, how much kinetic energy is lost?

For one body of mass m1 with velocity u colliding with another mass m2 at rest, the kinetic energy lost is (m1 m2 over m1 plus m2) times (u squared over 2). Maximum loss happens when the masses are equal — half the original KE is converted to heat or deformation.

What is the minimum speed at the top of a vertical circular loop?

For a body moving on the inside of a vertical loop of radius r, the minimum speed at the top is the square root of (g r). At this speed, gravity alone provides the centripetal force; the string or track tension is zero. Below this speed, the body falls off the loop.

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Work, Energy and PowerNEET Physics · Class 11 · NCERT Chapter 5

Introduction

Work done by a constant force

Sign of work

Work done by a variable force

Kinetic energy

The work-energy theorem

When to reach for it

Potential energy

Gravitational potential energy (near Earth)

Conservative force ↔ potential energy

Elastic (spring) potential energy

Conservation of mechanical energy

When energy is not conserved

Power

Collisions in one dimension

Perfectly elastic 1D collision

Perfectly inelastic 1D collision

Vertical circular motion

Speed at the top vs at the bottom

Minimum speed at the top

Worked NEET problems

Summary cheat sheet

Frequently asked questions

How many questions come from Work, Energy and Power in NEET 2027?

Why is Work, Energy and Power so important for NEET?

What is the work-energy theorem?

What is the formula for the energy stored in a spring?

What is the difference between average power and instantaneous power?

In a perfectly inelastic 1D collision, how much kinetic energy is lost?

What is the minimum speed at the top of a vertical circular loop?

Continue with the next chapter notes

Track Your NEET Score Across All 90 Chapters