NCERT Ch 6

Work, Energy & Power

The energy way of looking at motion · Class 11 Physics

1The big idea

The last two chapters tracked motion instant by instant with forces and accelerations. This chapter offers a shortcut. Instead of following every push at every moment, we keep a single running account — energy — and use the fact that it is only ever moved around or converted, never created or destroyed. Many problems that look horrible with force diagrams collapse to one line of energy bookkeeping.

The whole chapter hangs on one connecting idea: work is the bridge between force and energy. When a force does work on a body it deposits energy into it; when the body does work against a force it pays energy out. Get comfortable with that single sentence and kinetic energy, potential energy, power and collisions all fall into place as variations on it.

2What "work" really means

In everyday speech, holding a heavy suitcase still is hard "work". In physics it is zero work — nothing moves. The physicist's definition is narrow and precise: work is done only when a force produces a displacement along its own line.

Work done by a constant force W = F s cos θ

Here F is the force, s the displacement, and θ the angle between them. Only the component of the force in the direction of motion, F cos θ, counts. The unit is the joule (1 J = 1 N m). Because cos θ carries a sign, work comes in three flavours:

F θ F cos θ displacement s
Only the component of F along the motion, F cos θ, does work. The vertical part lifts nothing here, so it does none.

When the force varies (like a spring), we can no longer just multiply. Work then becomes the area under the force–displacement graph — add up F Δs over every tiny step. That area idea is what we use for the spring in a moment.

3Kinetic energy & the work–energy theorem

Kinetic energy is the energy a body has purely because it is moving:

Kinetic energy K = ½ m v2

Where does the ½ and the square come from? Watch a constant net force F act over a displacement s, taking the body from speed u to speed v. The work done is W = F s = m a s. Now borrow one equation of motion from kinematics, v2 = u2 + 2 a s, which rearranges to a s = (v2 − u2)/2. Substitute:

W = m · v2 − u22 = ½ m v2 − ½ m u2 = Kf − Ki

That result is the work–energy theorem: the net work done on a body equals the change in its kinetic energy. It is one of the most useful lines in mechanics — it turns "find the speed" problems into "find the work" problems, with no need to know the time taken. Notice the logic runs both ways: do positive work and the body speeds up; friction doing negative work is exactly why a sliding block eventually stops, its kinetic energy all cashed out as heat.

Remember this

Net work in = change in kinetic energy. Whenever a question gives you a force over a distance and asks for a speed (or the reverse), reach for W = ΔK before you reach for F = m a.

4Potential energy — stored work

Some work does not vanish into speed — it gets stored, ready to be handed back. Lift a book slowly and steadily and its kinetic energy barely changes, yet you clearly did work. That work is now held as potential energy: energy of position or configuration, waiting to be released.

Gravitational potential energy

To raise a mass m through a height h at steady speed you must push up with a force equal to its weight m g, through a distance h. The work you deposit is

Gravitational PE (near Earth) U = m g h

Only changes in height matter, so you are free to call any convenient level "zero" — the floor, the tabletop, sea level. What is physical is the difference m g Δh the body can pay back on the way down.

Elastic potential energy of a spring

A spring obeys Hooke's law: to stretch (or compress) it by x takes a force F = k x, growing steadily with the stretch (k is the spring constant, in N m−1). Because the force is not constant, we take the area under the F–x line. That area is a triangle of base x and height k x:

U = ½ × base × height = ½ · x · (k x) = ½ k x2

So a stretched or compressed spring stores ½ k x2, and hands it straight back as kinetic energy when released. The square means doubling the stretch quadruples the stored energy — a fact catapults and bows quietly rely on.

5Conservation of mechanical energy

Gravity and the spring force share a special property: the work they do depends only on the start and end points, not on the path taken. Such forces are called conservative, and for them we can define a potential energy at all. Now combine two things we already have. The work–energy theorem says the work by gravity equals the change in kinetic energy, Wgrav = ΔK. But by the very definition of potential energy, that same work is minus the change in PE, Wgrav = −ΔU (going down releases energy, going up stores it). Set them equal:

ΔK = −ΔU  ⟹  ΔK + ΔU = 0  ⟹  K + U = constant

This is conservation of mechanical energy: when only conservative forces act, the total E = K + U never changes. As a ball falls, PE turns into KE joule for joule; as a pendulum swings up, KE turns back into PE. The total sits still while the two halves trade places. Written out for two points,

Between any two points ½ m u2 + m g h1 = ½ m v2 + m g h2

Drop a ball from rest through a height h and all of m g h becomes ½ m v2, giving the tidy result v = √(2 g h) — the same answer kinematics gives, reached in one line and with no mention of time. (When friction or air resistance is present, mechanical energy is not conserved; the missing joules have simply become heat and sound. The grand total energy is still conserved — it has just left the mechanical account.)

◆ Activity — energy on a frictionless track

Set the release height h of a ball on a smooth curved track and let it go. Watch the two energy bars: as the ball drops, potential energy (blue) pours into kinetic energy (red), and climbs back the other side. Their sum never changes — that horizontal ceiling is the fixed total E = m g h. At the very bottom all of it is kinetic, so the speed is v = √(2 g h).

Worked example

A 0.20 kg ball is released from rest at the top of a smooth track 1.8 m high. Take g = 10 m s−2. (a) How fast is it moving at the bottom? (b) It then compresses a spring of constant k = 400 N m−1. How far does the spring compress before the ball stops?

(a) The track is smooth, so mechanical energy is conserved. All the potential energy at the top becomes kinetic energy at the bottom:

m g h = ½ m v2  ⟹  v = √(2 g h) = √(2 × 10 × 1.8) = √36 = 6 m s−1. (The mass cancels — every object would arrive at the same speed.)

(b) Now the kinetic energy is stored in the spring. Set ½ m v2 = ½ k x2:

x = v √(mk) = 6 × √(0.20400) = 6 × √0.0005 = 6 × 0.02236 ≈ 0.13 m.

Two energy conversions, gravitational → kinetic → elastic, chained through one conserved total. No forces, no accelerations, no time.

6Power — how fast work is done

Energy tells you how much; power tells you how fast. Two cranes may both lift a load one metre, doing the same work, but the one that does it in half the time is twice as powerful. Power is work per unit time:

Average power P = Wt

The unit is the watt (1 W = 1 J s−1). For an instantaneous value, take work over a vanishingly short time. Since a small piece of work is ΔW = F Δs, dividing by Δt gives F × (Δs/Δt) = F v. Including the angle between force and velocity:

Instantaneous power P = F v cos θ = F · v

This is why a car engine of fixed power can pull hard at low speed but only gently at high speed — F and v trade off to keep P constant. A handy commercial unit is the kilowatt-hour: a power of 1 kW running for 1 hour, which is energy (1 kW h = 3.6 × 106 J), not power — the "unit" on your electricity bill.

7Collisions — elastic & inelastic

When two bodies collide, the forces are violent and brief, so external forces (like gravity) hardly matter during the impact. That gives us a rock-solid anchor:

Remember this

Momentum is conserved in every collision — elastic, inelastic, sticky, explosive, all of them. Kinetic energy is conserved only in a perfectly elastic collision.

We sort collisions by what happens to the kinetic energy:

One-dimensional elastic collision

Write both conservation laws for masses m1, m2 hitting head-on with speeds u1, u2. Momentum: m1 u1 + m2 u2 = m1 v1 + m2 v2. Kinetic energy: ½ m1 u12 + ½ m2 u22 = ½ m1 v12 + ½ m2 v22. Solving the pair gives the final velocities:

v1 = m1 − m2m1 + m2 u1 + 2 m2m1 + m2 u2   ·   v2 = 2 m1m1 + m2 u1 + m2 − m1m1 + m2 u2

The one case worth memorising: equal masses. Put m1 = m2 and the first fractions vanish while the second become 1, leaving v1 = u2 and v2 = u1. The two bodies simply exchange velocities. A moving ball striking an identical stationary one stops dead and sends the other off at its full speed — exactly the click of a Newton's cradle or a clean pool shot.

Coefficient of restitution

Real collisions live between the elastic and perfectly-inelastic extremes. The coefficient of restitution e measures how "bouncy" a collision is — the ratio of the speed of separation to the speed of approach:

e = speed of separationspeed of approach = v2 − v1u1 − u2

e = 1 for a perfectly elastic collision (they separate as fast as they approached), e = 0 for a perfectly inelastic one (they don't separate at all — they stick), and 0 < e < 1 for everything real. A quick way to measure it: drop a ball from height h and let it rebound to height h′; then e = √(h′/h).

8Common confusions to clear up

9Check yourself

Class 11 Physics · Work, Energy & Power · aligned to NCERT Chapter 6 · SmartStudy.School