Chapter 3 of 18

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JEE Main Physics★ ~1 question (4 marks) in JEE Main, most often a collision or an energy-conservation problem with friction

Work, Energy and Power

Energy methods bypass the path. A problem that is intractable with forces collapses to one line — provided you know which conservation law is actually available.

⏱ 21 min readMedium difficulty✓ Free · exam-pattern aligned

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JEE Main PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Spot the three standard zero-work forces before writing any integral
2Apply $W_{n e t} = Δ K$ including friction, where energy conservation would not apply
3Combine the radial equation at the top with energy conservation to get the vertical-circle conditions, and know when $5 g r$ does not apply
4Read stable and unstable equilibria, turning points and bound motion straight off a $U$ – $x$ curve — a curve that exists only because the force is conservative
5Use $Δ K + Δ U = - f_{k} d$ with the path length, not the displacement
6Choose the conservation law a collision actually leaves available, in one dimension and in two, and quantify the kinetic energy lost

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Why this chapter matters in JEE Main

Energy relates the two endpoints of a motion without asking what happened in between, so a problem that is unsolvable with forces — a block on a curved track of unspecified shape — collapses to one line. Two decisions carry the chapter: whether the forces doing work are conservative, which decides if a potential energy exists at all, and which conservation law survives a collision. Momentum survives every collision; kinetic energy survives only elastic ones.

Before you start — revise these

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Scalar (dot) product of two vectors

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Integrating polynomials

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Newton's laws and free-body diagrams

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Uniform circular motion

Work, Energy and Power

A block slides down a curved, frictionless track of height $h$ , starting from rest. The track's shape is not given. Find its speed at the bottom.

With Newton's laws this is impossible — the normal force changes direction continuously and you were never told the shape.

With energy it is one line. The normal force is always perpendicular to the motion, so it does no work. Only gravity does work:

$\frac{1}{2} m v^{2} = m g h \Rightarrow v = 2 g h$

The shape never mattered.

That is the whole reason this chapter exists: energy relates the endpoints without asking what happened in between.

1. Work by a constant force

$W = F \cdot d = F d cos θ$

A scalar product $\Rightarrow$ work is a scalar with a sign but no direction.

$θ$	Work
$< 90°$	positive
$= 90°$	zero
$> 90°$	negative

Three zero-work cases worth memorising:

Centripetal force in uniform circular motion — always perpendicular to $v$
Normal force on a body sliding along a surface
Any force on a stationary body, however large

Trap. Work is done by a named force, never "by a body". A question asking for "the work done" without naming the force is testing whether you noticed.

Illustration 1

A 10 kg crate is dragged 4 m across level ground by a 50 N force at $37°$ above the horizontal. $μ_{k} = 0.2$ , $g = 10$ . Find the work done by each force on it.

Vertical equilibrium first — and the pull has an upward component, so $N \neq = m g$ :

$N + 50 sin 37° = m g \Rightarrow N = 100 - 30 = 70 N$ , so $f_{k} = 0.2 (70) = 14 N$

Force	$θ$ with displacement	Work
Applied 50 N	$37°$	$50 (0.8) (4) = + 160$ J
Friction 14 N	$180°$	$- 14 (4) = - 56$ J
Gravity	$90°$	$0$
Normal	$90°$	$0$

$W_{n e t} = 160 - 56 = 104$ J.

The point: using $N = m g = 100$ N would have given $f_{k} = 20$ N and the wrong answer. A force with a vertical component always changes the normal force.

2. Work by a variable force

$W = \int_{x_{1}}^{x_{2}} F d x$

Graphically: area under the $F$ – $x$ graph, with area below the axis negative.

For a spring, $F = - k x$ , so the work done by the spring stretching from 0 to $x$ is

$W_{s p r in g} = - \frac{1}{2} k x^{2}$

The minus sign is real — the spring opposes the stretch. The external agent does $+ \frac{1}{2} k x^{2}$ .

Work goes as $x^{2}$ . Stretching from 2 cm to 4 cm takes three times the work of 0 to 2 cm, not twice.

Illustration 2

Read the work done from $x = 0$ to $x = 6$ m off the graph above.

Split it into pieces whose areas you can name:

$0$ to $2$ : rectangle, $10 \times 2 = + 20$ J
$2$ to $4$ : triangle, $\frac{1}{2} (2) (10) = + 10$ J
$4$ to $6$ : triangle below the axis, $- \frac{1}{2} (2) (10) = - 10$ J

$W = 20 + 10 - 10 = + 20$ J.

Read it back: the particle gains kinetic energy up to $x = 4$ m, then loses some of it. It ends with exactly the kinetic energy it had at $x = 2$ m.

Illustration 3

A spring of force constant 200 N/m is stretched from 2 cm to 4 cm. Find the work the external agent must do.

$W = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2}) = 100 (0.0 4^{2} - 0.0 2^{2}) = 100 (0.0012) = 0.12 J$

Stretching the same spring from 0 to 2 cm costs $100 (0.0004) = 0.04$ J.

Three times the work for the same 2 cm — the quadratic in action.

3. Kinetic energy and the work–energy theorem

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

$W_{n e t} = Δ K = K_{f} - K_{i}$

The power is in the word net. Include every force — gravity, friction, tension, applied — and the path drops out.

Unlike energy conservation, this theorem holds even with friction. Friction just contributes negative work.

It is a scalar equation, so it gives one equation no matter how many dimensions the motion has. Convenient, and also its limitation.

The form $K = p^{2} /2 m$ is the one to reach for in collision problems, where one of $K$ and $p$ is given and the other is wanted.

Illustration 4

A bullet loses half its speed passing through one plank. How many more identical planks will it pass through before stopping?

Speed halved $\Rightarrow$ kinetic energy quartered. So one plank removes $\frac{3}{4} K_{0}$ and leaves $\frac{1}{4} K_{0}$ .

Each plank is identical, so each removes the same energy $\frac{3}{4} K_{0}$ :

$n = \frac{\frac{1}{4} K _{0}}{\frac{3}{4} K _{0}} = \frac{1}{3}$

Not one more. It buries itself one-third of the way into the next plank.

The trap: "loses half its speed" tempts you into "so one more plank". Energy, not speed, is what the plank removes.

4. Conservative forces and potential energy

Conservative — work around any closed path is zero; equivalently, work between two points is route-independent.

Conservative	Not conservative
Gravity, spring, electrostatic	Friction, air resistance, viscous drag

Only for a conservative force can a potential energy exist:

$Δ U = - W_{co n ser v a t i v e}, F = - \frac{d U}{d x}$

The minus sign encodes the trade: when gravity does positive work on a falling body, $U$ drops by exactly that much.

Friction fails decisively — slide a block round a closed loop and it comes back with less energy, so no potential function can exist.

Reading a potential-energy curve

A $U$ – $x$ graph contains the entire dynamics of a 1-D conservative system, and JEE Main asks you to read it directly.

Force $=$ negative slope $\Rightarrow$ the body is pushed downhill on the graph.
Zero slope $\Rightarrow$ equilibrium. Minimum = stable (restoring force), maximum = unstable, flat = neutral.
Total energy $E$ is a horizontal line. Since $K = E - U \geq 0$ , motion is confined to where the curve lies below the line.
Where the line meets the curve: turning points, speed momentarily zero.
Trapped between two turning points = bound. Only one turning point = unbound — which is exactly the escape condition in gravitation.

Illustration 5

A particle moves under $U (x) = \frac{x ^{3}}{3} - 4 x$ (SI units). Locate the equilibrium positions and classify them.

$F = - \frac{d U}{d x} = - (x^{2} - 4) = 4 - x^{2}$

$F = 0 \Rightarrow x = \pm 2$ m.

$\frac{d ^{2} U}{d x ^{2}} = 2 x$ :

$x$	$d^{2} U / d x^{2}$	Nature
$+ 2$	$+ 4$ (minimum)	stable
$- 2$	$- 4$ (maximum)	unstable

Verify with the force directly: at $x = 2.1$ , $F = 4 - 4.41 = - 0.41$ N, pushing the particle back toward $x = 2$ . Restoring, so stable.

5. Conservation of mechanical energy

$K + U = constant$

The condition is on the forces that do work, not the forces present. A normal force can act throughout without breaking conservation, because it does none.

Near the Earth: $U = m g h$ , with the reference level free — only differences matter. Spring: $U = \frac{1}{2} k x^{2}$ from natural length.

With non-conservative forces:

$Δ K + Δ U = W_{n c}$

For friction $W_{n c} = - f_{k} d$ , where $d$ is the actual path length, not the displacement. This is the one place in mechanics where distance, not displacement, is required.

Illustration 6

A 2 kg block slides 5 m down a $30°$ incline from rest, $μ_{k} = 0.2$ , $g = 10$ . Find its speed at the bottom.

Height dropped: $h = 5 sin 30° = 2.5 m$

$N = m g cos 30° = 2 (10) (0.866) = 17.3 N$ , so $f_{k} = 0.2 (17.3) = 3.46 N$

$Δ K + Δ U = W_{n c}$ :

$\frac{1}{2} (2) v^{2} - 0 + (0 - 2 (10) (2.5)) = - 3.46 (5)$

$v^{2} - 50 = - 17.3 \Rightarrow v^{2} = 32.7 \Rightarrow v \approx 5.7 m/s$

Check: frictionless would give $2 (10) (2.5) = 7.07$ m/s. Friction reduced it, as it must.

Illustration 7

A 0.5 kg block sliding at 4 m/s on a rough floor ( $μ_{k} = 0.25$ ) runs into a spring of $k = 100$ N/m. Find the maximum compression. $g = 10$ .

At maximum compression the block is momentarily at rest, so all its kinetic energy has gone into the spring and into friction over the same distance $x$ :

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2} + μ_{k} m g x$

$4 = 50 x^{2} + 1.25 x \Rightarrow 50 x^{2} + 1.25 x - 4 = 0$

$x = \frac{- 1.25 + 1.5625 + 800}{100} = \frac{- 1.25 + 28.31}{100} \approx 0.27 m$

Check: with no friction, $\frac{1}{2} (100) x^{2} = 4$ gives $x = 0.283$ m. Friction shortened the compression slightly, as it must.

6. Motion in a vertical circle

A body on a string swung in a vertical plane is the standard test of whether you can use energy and Newton's second law together. Energy alone will not do it.

Two equations, applied at the same instant:

Energy, between the lowest point and a height $h$ : $v^{2} = v_{b}^{2} - 2 g h$
Radial Newton, at that point: net inward force $= \frac{m v ^{2}}{r}$

The critical condition is at the top, where gravity points along the required centripetal direction and the string can only pull:

$T + m g = \frac{m v _{t}^{2}}{r}$ , and a string needs $T \geq 0$

$v_{t}^{min} = g r$

Now carry that down with energy, through a height $2 r$ :

$v_{b}^{2} = v_{t}^{2} + 4 g r = g r + 4 g r \Rightarrow v_{b}^{min} = 5 g r$

At the bottom, $T_{b} - m g = \frac{m v _{b}^{2}}{r} = 5 m g$ , so $T_{b} = 6 m g$ while $T_{t} = 0$ .

The 6mg result is general. For any speed at which the circle is completed, $T_{b} - T_{t} = 6 m g$ exactly — the $2 r$ height difference contributes $4 m g$ through the energy equation and the reversal of gravity's direction contributes $2 m g$ .

Three regimes for a string, by the speed at the lowest point:

$v_{b}$	What happens
$\geq 5 g r$	completes the circle
between $2 g r$ and $5 g r$	string goes slack above the horizontal; the body leaves the circle and becomes a projectile
$\leq 2 g r$	oscillates below the horizontal like a pendulum, string stays taut

Trap. $5 g r$ applies to a string, or the inside of a track — anything that can only pull or push inward. A rod can also push outward, so it supports the body at the top at zero speed: the condition becomes $v_{t} \geq 0$ and hence $v_{b} \geq 4 g r = 2 g r$ .

Illustration 8

A 200 g stone on a 1 m string is whirled in a vertical circle, $g = 10$ . Find the minimum speed at the top, the corresponding speed at the bottom, and the tension at the bottom then.

Top: $v_{t} = g r = 10 = 3.16$ m/s

Bottom: $v_{b} = 5 g r = 50 = 7.07$ m/s

Tension: $T_{b} = 6 m g = 6 (0.2) (10) = 12 N$

Check: $T_{b} - T_{t} = 12 - 0 = 12$ N, and $6 m g = 12$ N. The general result holds.

7. Power

$P_{a v g} = \frac{W}{t}, P_{in s t} = \frac{d W}{d t} = F \cdot v$

At constant power, $F$ and $v$ are inversely related — which is why a car's acceleration falls as it speeds up even at full throttle, and why top speed is reached exactly when engine power is fully absorbed by drag.

Setting $P = F v$ with drag $F = k v^{2}$ gives $v_{ma x} = (P / k)^{1/3}$ : eight times the engine power for twice the top speed.

Illustration 9

A pump raises 600 kg of water per minute through 20 m and discharges it at 5 m/s. Find the power delivered. $g = 10$ .

Per second the pump handles $600/60 = 10$ kg.

Term	Rate
Potential energy	$10 (10) (20) = 2000$ W
Kinetic energy	$\frac{1}{2} (10) (5^{2}) = 125$ W

$P = 2125 W \approx 2.1 kW$

The trap: the water leaves moving, so the kinetic term is real. Dropping it loses 125 W — and dropping it is the commonest error in this question type.

8. Collisions

Momentum is conserved in every collision. Kinetic energy only in elastic ones. Knowing which law is available is the entire skill.

Momentum survives because the internal forces are third-law pairs and cancel, provided external forces are negligible during the brief contact.

$e = \frac{relative velocity of separation}{relative velocity of approach}$

$e$	Type	$K$
1	Perfectly elastic	conserved
$0 < e < 1$	Real collisions	partly lost
0	Perfectly inelastic	maximum loss consistent with momentum

For a 1-D elastic collision, solving momentum and energy together:

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

Two special cases, worth knowing cold:

Equal masses exchange velocities exactly. Set $m_{1} = m_{2}$ and the first term vanishes.
A light body hitting a heavy stationary one rebounds at nearly its original speed.

Perfectly inelastic: they move together at $\frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ .

Illustration 10

A 2 kg ball at 6 m/s hits a stationary 4 kg ball head-on and elastically. Find both final velocities.

$v_{1} = \frac{2 - 4}{6} (6) = - 2 m/s$ — it rebounds.

$v_{2} = \frac{2 ( 2 )}{6} (6) = 4 m/s$

Check momentum: $2 (6) = 12$ before; $2 (- 2) + 4 (4) = 12$ after.

Check energy: $\frac{1}{2} (2) (36) = 36$ J before; $\frac{1}{2} (2) (4) + \frac{1}{2} (4) (16) = 4 + 32 = 36$ J after. Elastic confirmed.

How much energy a perfectly inelastic collision loses

$Δ K = \frac{1}{2} (\frac{m _{1} m _{2}}{m _{1} + m _{2}}) (u_{1} - u_{2})^{2}$

The bracket is the reduced mass. Read the formula: the loss depends only on the relative velocity, so two bodies moving together at any common speed lose nothing.

With the target at rest, the fraction of kinetic energy lost is simply $\frac{m _{2}}{m _{1} + m _{2}}$ .

A heavy projectile on a light target loses almost nothing; a light projectile on a heavy target loses almost everything. This is why a neutron is slowed by hydrogen and not by lead.

Illustration 11

A 2 kg block at 5 m/s strikes a stationary 3 kg block and they stick together. Find the common velocity and the percentage of kinetic energy lost.

$v = \frac{2 ( 5 )}{5} = 2 m/s$

$K_{i} = \frac{1}{2} (2) (25) = 25$ J, $K_{f} = \frac{1}{2} (5) (4) = 10$ J, lost $= 15$ J $= 60%$

Check with the formula: $\frac{m _{2}}{m _{1} + m _{2}} = \frac{3}{5} = 60%$ . Agrees.

A ball bouncing on the floor

Treat the floor as a body of infinite mass. Then $e$ compares rebound speed with impact speed, and since $h \propto v^{2}$ :

$h_{n} = e^{2 n} h, v_{n} = e^{n} v_{0}$

Summing the geometric series over all bounces:

Total distance: $h \frac{1 + e ^{2}}{1 - e ^{2}}$
Total time: $\frac{2 h}{g} \cdot \frac{1 + e}{1 - e}$

Both are finite even though the number of bounces is infinite — the ball comes to rest in a definite time.

Illustration 12

A ball dropped from 5 m rebounds to 1.8 m. Find $e$ , the height after the second bounce, and the total distance it travels.

$e = h_{1} / h = 1.8/5 = 0.36 = 0.6$

$h_{2} = e^{4} h = 0.1296 (5) = 0.648 m$

Total distance $= 5 (\frac{1 + 0.36}{1 - 0.36}) = 5 (\frac{1.36}{0.64}) = 10.625 m$

9. Collisions in two dimensions

Momentum is a vector, so it is conserved component by component — two equations instead of one. Kinetic energy stays a single scalar equation.

For two smooth spheres, resolve along two special directions at the moment of contact:

Direction	What is true
Line of impact (common normal, through both centres)	momentum conservation and the restitution equation apply
Common tangent (perpendicular to it)	no impulse acts, so each body keeps its own component unchanged

That second row is the whole technique: the tangential components simply pass through the collision untouched.

The right-angle result, in three lines. Equal masses, target at rest, perfectly elastic:

Momentum: $u = v_{1} + v_{2}$

Energy: $u^{2} = v_{1}^{2} + v_{2}^{2}$

Square the first: $u^{2} = v_{1}^{2} + v_{2}^{2} + 2 v_{1} \cdot v_{2}$

Comparing, $v_{1} \cdot v_{2} = 0$ , so the two velocities are perpendicular — whatever the impact angle was.

The right angle is a consequence of elasticity, not an assumption. If the collision is inelastic the angle closes to less than $90°$ , which is how bubble-chamber photographs reveal that a collision was not elastic.

Illustration 13

A ball at 10 m/s strikes an identical stationary ball elastically and moves off at $30°$ to its original direction. Find both final speeds and the second ball's direction.

By the right-angle result the second ball goes off at $60°$ on the other side. Now use components.

Along the original direction: $10 = v_{1} cos 30° + v_{2} cos 60°$

Perpendicular to it: $0 = v_{1} sin 30° - v_{2} sin 60°$

From the second: $0.5 v_{1} = 0.866 v_{2} \Rightarrow v_{1} = 1.732 v_{2}$

Substituting: $10 = 1.732 v_{2} (0.866) + 0.5 v_{2} = 1.5 v_{2} + 0.5 v_{2} = 2 v_{2}$

$v_{2} = 5 m/s, v_{1} = 8.66 m/s$

Check energy: $1 0^{2} = 100$ before; $8.6 6^{2} + 5^{2} = 75 + 25 = 100$ after. Elastic confirmed.

Summary

Energy relates endpoints without the path. That is why it beats forces when the geometry is unknown.
$W = F d cos θ$ , a signed scalar. Centripetal force, normal force on a sliding body, and any force on a stationary body all do zero work.
A pull with a vertical component changes $N$ , and therefore changes friction.
Variable force: $W = \int F d x$ , the signed area under $F$ – $x$ . Spring work goes as $x^{2}$ .
$W_{n e t} = Δ K$ holds even with friction — unlike energy conservation.
Potential energy exists only for conservative forces. $F = - d U / d x$ ; sign of $d^{2} U / d x^{2}$ decides stability.
On a $U$ – $x$ graph: minimum stable, maximum unstable, $K = E - U$ , turning points where they meet.
$Δ K + Δ U = W_{n c}$ , and for friction that is $- f_{k} d$ using path length.
Vertical circle, string: $v_{t o p} = g r$ , $v_{b o tt o m} = 5 g r$ , and $T_{b} - T_{t} = 6 m g$ always. A rod needs only $2 g r$ .
$P_{in s t} = F \cdot v$ . Constant power means $F$ falls as $v$ rises, giving $v_{ma x} \propto P^{1/3}$ .
Momentum: conserved in every collision. Kinetic energy: elastic only.
Equal masses in an elastic 1-D collision exchange velocities.
Perfectly inelastic loss depends on relative velocity; with the target at rest the fraction lost is $m_{2} / (m_{1} + m_{2})$ .
In 2-D, tangential components pass through unchanged. Equal masses, one at rest, elastic: they separate at $90°$ .

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

Work

W = F \cdot d = F d cos θ, W = \int_{x_{1}}^{x_{2}} F d x

A signed scalar. Zero when $\theta=90°$ — which covers the centripetal force, the normal force on a sliding body, and any force on a stationary body. For a variable force it is the area under the $F$–$x$ graph.

Kinetic energy

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

The $p^{2}/2m$ form is the one to reach for in collision questions, where one of $K$ and $p$ is given and the other is wanted.

Work-energy theorem

W_{n e t} = Δ K = K_{f} - K_{i}

The word *net* is the whole point — include every force, and the path drops out. Unlike energy conservation this holds **even with friction**, which simply contributes negative work. One scalar equation regardless of dimension.

Spring: force, work and energy

F = - k x, W_{s p r in g} = - \frac{1}{2} k x^{2}, U = \frac{1}{2} k x^{2}

The minus on the work is real — the spring opposes the stretch while the external agent does $+\tfrac12kx^{2}$. Work goes as $x^{2}$, so 2 cm to 4 cm costs **three times** as much as 0 to 2 cm.

Potential energy and force

Δ U = - W_{co n s}, F = - \frac{d U}{d x}

Defined only for conservative forces (gravity, spring, electrostatic) — never for friction, which is path-dependent. On a $U$–$x$ graph the force is the negative slope, so a minimum is stable equilibrium and a maximum is unstable.

Mechanical energy

K + U = constant, Δ K + Δ U = W_{n c}

The condition is on the forces that *do work*, not those present — a normal force may act throughout without breaking conservation. For friction $W_{nc}=-f_kd$ with $d$ the **actual path length**, the one place distance beats displacement.

Power

P_{a v g} = \frac{W}{t}, P_{in s t} = \frac{d W}{d t} = F \cdot v

At constant power $F$ and $v$ are inversely related, which is why acceleration falls as a car speeds up, and why top speed is where engine power exactly matches drag.

Collisions

e = \frac{v _{se p}}{v _{a pp}}, v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} + \frac{2 m _{2}}{m _{1} + m _{2}} u_{2}

Momentum is conserved in **every** collision; kinetic energy only when $e=1$. Equal masses in a 1-D elastic collision exchange velocities exactly. Perfectly inelastic ($e=0$) bodies move together at $\dfrac{m_1u_1+m_2u_2}{m_1+m_2}$.

Vertical circle (string or inside of a track)

v_{t o p} = g r, v_{b o t} = 5 g r, T_{b o t} - T_{t o p} = 6 m g

The critical point is the **top**, where a string can only pull: $T\ge0$ forces $v_{top}\ge\sqrt{gr}$, and energy over the height $2r$ gives $\sqrt{5gr}$ at the bottom. The $6mg$ gap holds for *any* speed that completes the circle. A **rod** can push outward, so it needs only $v_{top}\ge0$, i.e. $v_{bot}\ge2\sqrt{gr}$.

Energy lost in a perfectly inelastic collision

Δ K = \frac{1}{2} (\frac{m _{1} m _{2}}{m _{1} + m _{2}}) (u_{1} - u_{2})^{2}

The bracket is the **reduced mass**. The loss depends only on the *relative* velocity, so bodies already moving together lose nothing. With the target at rest the fraction lost is $\dfrac{m_2}{m_1+m_2}$ — near zero for a heavy projectile on a light target, near one for the reverse.

Successive bounces off the floor

h_{n} = e^{2 n} h, d_{t o t a l} = h \frac{1 + e ^{2}}{1 - e ^{2}}, t_{t o t a l} = \frac{2 h}{g} \cdot \frac{1 + e}{1 - e}

Restitution multiplies the *speed* by $e$, and $h\propto v^{2}$, so the **height** multiplies by $e^{2}$. Both totals are finite despite infinitely many bounces. Check the limits: $e=0$ gives $d=h$, and $e\to1$ makes both diverge.

Oblique collision of two smooth bodies

u = v_{1} + v_{2} and u^{2} = v_{1}^{2} + v_{2}^{2} \Rightarrow v_{1} \cdot v_{2} = 0

Resolve along the **line of impact** (momentum plus restitution apply) and along the **common tangent** (no impulse, so each body keeps its own component). For equal masses with one at rest and $e=1$, squaring the momentum equation and comparing with the energy equation forces the two to separate at exactly $90°$.

Top speed at constant engine power

P = F v, F_{d r a g} = k v^{2} \Rightarrow v_{ma x} = (\frac{P}{k})^{1/3}

Top speed is reached when the driving force has fallen to equal the drag. The cube root is brutal: **eight times the engine power buys only twice the top speed**.

JEE Main test series

Ready to be tested on this?

Work, Energy and Power shows up in every JEE Main paper — sit a timed one and find out where you actually stand.

~1Est. Qs

Highpriority

Full JEE Main mocks, real clock
Physics-only sectionals & chapter tests
All-India rank + topic-level analysis

Free forever · solver-verified questions

⚠️

Traps JEE Main sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Using

- f_{k} \times

displacement for the work done by friction

✓ Friction acts along the actual path, so its work is

- f_{k} d

with

d

the path length. On a closed loop the displacement is zero but the friction work is not.

Why it happens: Every other work calculation in the chapter uses displacement, so it becomes automatic.

WATCH OUT

✗ Applying

K + U =

constant when friction is present

✓ Use

Δ K + Δ U = W_{n c}

instead. The work–energy theorem

W_{n e t} = Δ K

also still works — only mechanical energy conservation fails.

Why it happens: The two statements look almost identical, and conservation is the one drilled hardest.

WATCH OUT

✗ Assuming kinetic energy is conserved in every collision

✓ Momentum is conserved always; kinetic energy only when the collision is elastic. If the question says the bodies stick together,

K

is definitely lost.

Why it happens: Both laws are taught together in the same section, so they get applied together.

WATCH OUT

✗ Averaging a variable force and multiplying by the distance

✓ That works only if

F

is linear in

x

. Otherwise integrate: for

F = 3 x^{2} + 2 x

from 0 to 2 the answer is 12 J, while the midpoint force of 5 N would give 10 J.

Why it happens: The linear spring case is met first and does work that way.

WATCH OUT

✗ Reading a maximum on a

U

–

x

curve as stable equilibrium

✓ Both maxima and minima have zero slope, but only a minimum gives a restoring force. At a maximum a small displacement pushes the body further away.

Why it happens: Both are called equilibrium points and look alike on the graph.

WATCH OUT

✗ Saying the normal force breaks energy conservation on a curved track

✓ It does no work, because it is always perpendicular to the velocity. Mechanical energy is conserved on any frictionless track whatever its shape — which is what makes those problems solvable.

Why it happens: The normal force is present and changing, so it looks like it must be doing something.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Work, Energy and Power?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min worth ~4 marks in JEE Main exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Energy relates endpoints without the path — that is why it beats forces when the geometry is unknown.
• $W = F d cos θ$ , a signed scalar. Zero work: centripetal force, normal force on a sliding body, any force on a stationary body.
•Variable force: $W = \int F d x$ , the area under $F$ – $x$ . Averaging the force only works if $F$ is linear.
• $W_{n e t} = Δ K$ holds even with friction. Mechanical energy conservation does not.
•Potential energy exists only for conservative forces. $F = - d U / d x$ .
• $U$ – $x$ curve: minimum stable, maximum unstable, $K = E - U$ , turning points where $E$ meets $U$ .
• $Δ K + Δ U = W_{n c}$ , and friction gives $- f_{k} d$ using the path length.
•Spring work goes as $x^{2}$ : 2 cm to 4 cm costs three times 0 to 2 cm.
• $P_{in s t} = F \cdot v$ . At constant power, force falls as speed rises.
•Momentum conserved in every collision; kinetic energy only if elastic. Equal masses exchange velocities.
•Vertical circle on a string: $g r$ at the top, $5 g r$ at the bottom, $T_{b o t} - T_{t o p} = 6 m g$ always. A rod needs only $2 g r$ .
•In 2-D, tangential components pass through a collision unchanged. Equal masses, one at rest, elastic: they separate at exactly $90°$ .

JEE Main question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: ~1 question (4 marks) of the 100-mark Physics section

Question style	Marks each	Typical count	What it tests
Work and the work-energy theorem	4	1	Signed work, the zero-work cases, work by a variable force as an integral or an area, and $W_{net}=\Delta K$
Energy conservation	4	1	Conservative versus non-conservative forces, reading a $U$–$x$ curve, the friction correction term, and the vertical circle
Collisions	4	1	Coefficient of restitution, 1-D elastic and perfectly inelastic collisions, and the successive-bounce series
Power	4	1	Average versus instantaneous power, $P=\vec{F}\cdot\vec{v}$, and pump or engine problems with both kinetic and potential terms

Prep strategy

For every problem, write down which forces do work and which do not before starting. That single habit prevents most errors in the chapter.
Derive the vertical-circle result once from the tension condition plus energy conservation, rather than memorising $\sqrt{5gL}$.
Practise collision questions by checking both conservation laws at the end. If kinetic energy fails to balance in a problem stated as elastic, the arithmetic is wrong.

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Before choosing a method, ask whether the path is known. If it is not, energy is almost certainly the intended route.
List which forces do work before writing anything. Ruling out the normal force and the tension usually leaves one or two terms.
If friction appears, switch from $K + U =$ constant to $Δ K + Δ U = - f_{k} d$ and use the path length.
In collisions, write momentum conservation first — it is always available. Only add energy conservation once the question says elastic.
For a $U$ – $x$ graph question, draw the total-energy line immediately. Turning points and the allowed region then read straight off the picture.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Regenerative braking recovers a vehicle's kinetic energy …

Regenerative braking recovers a vehicle's kinetic energy as electrical energy instead of dumping it into brake pads as heat, which is the work-energy theorem run in reverse.

Crumple zones and helmet liners are engineered to increas…

Crumple zones and helmet liners are engineered to increase the collision time, lowering the peak force for the same change in momentum.

Hydroelectric output is exactly $\overset{m}{˙} g h$

Hydroelectric output is exactly $\overset{m}{˙} g h$ — the mass flow rate times $g$ times the head — which is the pump problem in this chapter with the sign reversed.

Where else this topic is tested

Prepare once, score in every exam that asks it.

JEE Main

JEE Advanced

NEET UG

CBSE Class 11 Boards

BITSAT

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Because it is perpendicular to the surface, and the velocity is always along the surface. Work is the dot product of force and displacement, and a dot product of perpendicular vectors is zero at every instant — so the total is zero however the track bends. This is exactly what makes the classic problem solvable: a block released from height h on a frictionless track of completely unspecified shape arrives at the bottom with the speed root 2gh, because gravity is the only force doing any work and gravity only cares about the height dropped.

The work-energy theorem, net work equals change in kinetic energy, is always true. Mechanical energy conservation is a special case that additionally requires every force doing work to be conservative. So if friction, air resistance or an external applied force is doing work, conservation fails but the theorem does not. The safe habit is to start from the theorem and only specialise to conservation once you have checked that nothing non-conservative is doing work. Note the condition is on forces that do work, not forces present — a normal force can act throughout and change nothing.

Because friction always points opposite to the instantaneous direction of motion, so it always does negative work no matter which way the body goes. Over a path that doubles back, the two contributions add rather than cancel. Gravity is the opposite case: it points in a fixed direction, so going up cancels coming down and only the net height change matters. The cleanest test is a closed loop. The displacement is zero, so gravity does zero work over the loop, but a block dragged round it clearly loses energy to friction.

Momentum conservation follows from the third law: the forces the two bodies exert on each other are equal and opposite at every instant, so the impulses they deliver cancel exactly and the total momentum cannot change. That argument says nothing about energy. Kinetic energy can be converted into deformation, heat and sound during the contact, and it usually is. Only if the bodies deform elastically and spring fully back is all of it returned, which is what elastic means. That is why the coefficient of restitution measures elasticity and not momentum.

Because power is the product of force and velocity, so as the velocity rises the force can fall while keeping the product fixed. A car engine at full throttle roughly delivers constant power, so its driving force decreases as it speeds up, and its acceleration decreases with it. The top speed is reached exactly when the driving force has fallen to equal the resistive force, at which point acceleration is zero. This is also why a car accelerates fastest in low gear, where the gearing trades speed for force.

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Sources and How This Chapter Was CheckedSyllabus scope, what was derived rather than quoted, and how every answer here was checked.

Scope follows the NTA JEE Main syllabus for 2026 (Unit 4, Work, Energy and Power): work by constant and variable forces, kinetic and potential energy, the work–energy theorem, power, conservative and non-conservative forces, conservation of mechanical energy, motion in a vertical circle, and elastic and inelastic collisions in one and two dimensions.

Results were derived rather than quoted: spring work by integrating $F = - k x$ ; the $U$ – $x$ reading rules from $F = - d U / d x$ ; the friction term $- f_{k} d$ from work done along the actual path; the vertical-circle conditions by combining the radial equation at the top with energy over the height $2 r$ ; the elastic-collision velocity formula by solving momentum and kinetic-energy conservation simultaneously; and the right-angle result in two dimensions by squaring the vector momentum equation and comparing it with the energy equation.

Every illustration was checked against a second route — the incline result against the frictionless case, the spring compression against its frictionless value, the inelastic loss against the reduced-mass formula, the vertical-circle tensions against the general $T_{b} - T_{t} = 6 m g$ identity, and both collision problems against momentum and kinetic-energy conservation independently. The illustrations are teaching problems written for this chapter, not previous-year questions, and are not labelled as such.

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Work, Energy and Power

By the end of this chapter you'll be able to…

Before you start — revise these

Work, Energy and Power

1. Work by a constant force

Illustration 1

2. Work by a variable force

Illustration 2

Illustration 3

3. Kinetic energy and the work–energy theorem

Illustration 4

4. Conservative forces and potential energy

Reading a potential-energy curve

Illustration 5

5. Conservation of mechanical energy

Illustration 6

Illustration 7

6. Motion in a vertical circle

Illustration 8

7. Power

Illustration 9

8. Collisions

Illustration 10

How much energy a perfectly inelastic collision loses

Illustration 11

A ball bouncing on the floor

Illustration 12

9. Collisions in two dimensions

Illustration 13

Summary

Key formulas & results

Traps JEE Main sets — and how to dodge them

Exam-pattern practice

Are you exam-ready for Work, Energy and Power?

5-minute revision

JEE Main question blueprint

Exam-hall strategy

Beyond the exam

Where else this topic is tested

📝 My notes

Questions aspirants ask

Why does the normal force never do work, even on a curved track?

When can I use energy conservation and when must I use the work-energy theorem?

Why is friction's work computed with distance rather than displacement?

In a collision, why is momentum always conserved but kinetic energy not?

How can power be constant while force keeps changing?

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