Work, Energy and Power — Theorems, Conservation, Collisions

The Energy Framework

Work, energy, and power give us an alternative to Newton’s laws for solving mechanics problems. Instead of tracking forces and accelerations, we track energy — how it’s transferred (work), stored (potential energy), and used (power). The work-energy theorem and conservation of energy are two of the most powerful tools in physics.

This chapter carries 6-8 marks in CBSE Class 11. JEE Main tests 1-2 questions every session, and collision problems are an Advanced favourite.

graph TD
    A[Mechanics Problem] --> B{Energy approach useful?}
    B -->|Variable force| C[Work = ∫F·dx]
    B -->|Conservation scenario| D[KE + PE = constant]
    B -->|Power/efficiency| E[P = Fv or P = W/t]
    B -->|Collision| F{Type?}
    F -->|Elastic| G[KE + momentum conserved]
    F -->|Inelastic| H[Only momentum conserved]
    F -->|Perfectly inelastic| I[Bodies stick together]
    D --> J[Set up: initial E = final E]
    J --> K[Solve for unknown]

Essential Formulas

W = \vec{F} \cdot \vec{d} = Fd\cos\theta

For variable force: $W = \int_{x_1}^{x_2} F(x)\, dx$

Work by gravity: $W = mgh$ (downward displacement). Work by friction: $W = -\mu mg \cdot d$ (always negative).

W_{net} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

The net work done by ALL forces equals the change in kinetic energy.

If only conservative forces (gravity, spring) act:

KE_i + PE_i = KE_f + PE_f

Potential energy: Gravity $= mgh$ , Spring $= \frac{1}{2}kx^2$ .

P = \frac{W}{t} = \vec{F} \cdot \vec{v} = Fv\cos\theta

Unit: watt (W). 1 HP = 746 W.

Elastic: $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Coefficient of restitution: $e = \frac{v_2 - v_1}{u_1 - u_2}$ ( $e = 1$ elastic, $e = 0$ perfectly inelastic)

Solved Examples

Example 1 (Easy — CBSE)

A 2 kg block is lifted 5 m vertically. Find work done by gravity and by the lifting force.

Work by gravity $= -mgh = -2(10)(5) = \mathbf{-100 \text{ J}}$ (opposes motion).

Work by lifting force $= +100$ J (assuming constant speed, so net work = 0).

Example 2 (Medium — JEE Main)

A block of mass 1 kg slides down a frictionless incline of height 5 m. Find its speed at the bottom.

$mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = \mathbf{10 \text{ m/s}}$

Example 3 (Hard — JEE Main)

A 2 kg ball moving at 6 m/s collides head-on with a stationary 4 kg ball. If the collision is elastic, find velocities after collision.

Using elastic collision formulas:

$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 = \frac{2-4}{6} \times 6 = -2$ m/s (bounces back)

$v_2 = \frac{2m_1}{m_1 + m_2}u_1 = \frac{4}{6} \times 6 = 4$ m/s

Common Mistakes to Avoid

Mistake 1 — Forgetting $\cos\theta$ in work formula. Work depends on the component of force along displacement. At $90°$ , work is zero (normal force does no work on circular motion).

Mistake 2 — Applying energy conservation when non-conservative forces act. If friction is present, $KE_i + PE_i \neq KE_f + PE_f$ . You must include work done by friction.

Mistake 3 — Confusing elastic with perfectly inelastic. In elastic, KE is conserved. In perfectly inelastic, bodies stick together and KE is NOT conserved (maximum KE loss).

Mistake 4 — Wrong sign for spring PE. Spring PE is always positive: $\frac{1}{2}kx^2$ . Whether stretched or compressed, it stores energy.

Practice Questions

Q1. A spring with $k = 200$ N/m is compressed 0.1 m. Find the PE stored.

$PE = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.01) = 1$ J.

Q2. A car of mass 1000 kg travels at 20 m/s. Find its KE.

$KE = \frac{1}{2}mv^2 = 500 \times 400 = 200000$ J $= 200$ kJ.

Q3. A pump lifts 200 kg of water per minute to a height of 10 m. Find the power.

$P = mgh/t = 200 \times 10 \times 10/60 = 333.3$ W.

Q4. A 5 kg ball is dropped from 20 m. Find speed at 5 m above ground.

$mg(20-5) = \frac{1}{2}mv^2$ . $v = \sqrt{2 \times 10 \times 15} = \sqrt{300} \approx 17.3$ m/s.

Q5. A bullet of 10 g moving at 400 m/s embeds in a 2 kg block. Find velocity after collision.

Momentum conservation: $0.01 \times 400 = 2.01 \times v$ . $v = 4/2.01 \approx 1.99$ m/s.

FAQs

Can work be negative?

Yes. When force opposes displacement ( $\theta > 90°$ ), work is negative. Friction always does negative work on a moving object.

What is the difference between conservative and non-conservative forces?

Conservative forces (gravity, spring) have path-independent work — they only depend on start and end positions. Non-conservative forces (friction, air resistance) are path-dependent.

Why is KE always positive?

$KE = \frac{1}{2}mv^2$ . Mass is positive, $v^2$ is positive. KE is zero only when velocity is zero.

What happens to KE lost in an inelastic collision?

It converts to heat, sound, deformation, and other non-mechanical forms of energy. Total energy is still conserved — just not mechanical energy.

Advanced Concepts

Work done by a variable force

When force varies with position, we must integrate. The work done by a spring force $F = -kx$ from position $x_1$ to $x_2$ :

W = \int_{x_1}^{x_2} (-kx)\, dx = -\frac{1}{2}k(x_2^2 - x_1^2)

If released from compression $x_0$ to natural length (0):

W = \frac{1}{2}kx_0^2

This energy converts entirely to kinetic energy (no friction).

Work-energy theorem — the full version

The work-energy theorem says $W_{net} = \Delta KE$ , where $W_{net}$ includes work by ALL forces — conservative and non-conservative.

If we separate conservative forces:

W_{non-conservative} = \Delta KE + \Delta PE = \Delta E_{mechanical}

For friction: $W_{friction} = -\mu mg \cdot d$ (always negative, always reduces mechanical energy).

A 5 kg block slides 10 m down a 30 $°$ rough incline ( $\mu = 0.2$ ). Find speed at the bottom.

Height lost: $h = 10\sin 30° = 5$ m

Work by gravity: $W_g = mgh = 5 \times 10 \times 5 = 250$ J

Work by friction: $W_f = -\mu mg\cos\theta \times d = -0.2 \times 5 \times 10 \times \cos 30° \times 10 = -86.6$ J

By work-energy theorem: $\frac{1}{2}mv^2 = 250 - 86.6 = 163.4$ J

$v = \sqrt{\frac{2 \times 163.4}{5}} = \sqrt{65.36} = \mathbf{8.08 \text{ m/s}}$

Elastic collision formulas (1D)

For two objects in a head-on elastic collision:

v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2}

v_2 = \frac{(m_2 - m_1)u_2 + 2m_1 u_1}{m_1 + m_2}

Special cases:

Equal masses ( $m_1 = m_2$ ): velocities exchange. $v_1 = u_2$ , $v_2 = u_1$ .
Heavy hits light at rest ( $m_1 \gg m_2$ , $u_2 = 0$ ): $v_1 \approx u_1$ , $v_2 \approx 2u_1$ .
Light hits heavy at rest ( $m_1 \ll m_2$ , $u_2 = 0$ ): $v_1 \approx -u_1$ (bounces back), $v_2 \approx 0$ .

Power applications

Maximum velocity on a level road: When engine power equals power lost to friction:

P = F_{friction} \times v_{max} = \mu mg \times v_{max}

v_{max} = \frac{P}{\mu mg}

Velocity of a car on a hill: At constant speed, the engine must overcome both gravity and friction:

P = (mg\sin\theta + \mu mg\cos\theta) \times v

JEE Main frequently asks: “A car of mass $m$ has power $P$ on a level road with friction coefficient $\mu$ . Find maximum speed.” This is a direct application: $v_{max} = P/(\mu mg)$ .

Percentage KE loss in perfectly inelastic collision

When two objects stick together after collision:

\text{Fractional KE loss} = \frac{m_2}{m_1 + m_2}

(where object 2 is initially at rest). The heavier the target relative to the projectile, the more KE is lost. This is why a bullet embedding in a block loses almost all its KE.

Additional Solved Examples

Example 4 (JEE Main): A 1 kg ball drops from 10 m onto a floor and bounces back to 6.4 m. Find the coefficient of restitution.

$e = \sqrt{\frac{h_{bounce}}{h_{drop}}} = \sqrt{\frac{6.4}{10}} = \sqrt{0.64} = \mathbf{0.8}$

Example 5 (JEE Advanced): A chain of mass $m$ and length $L$ lies on a table with $1/3$ hanging over the edge. Find the work needed to pull it entirely onto the table.

The centre of mass of the hanging portion is at depth $L/6$ below the table edge. Mass of hanging portion = $m/3$ .

$W = \frac{m}{3} \times g \times \frac{L}{6} = \frac{mgL}{18}$

Additional Practice Questions

Q6. A bullet of mass 20 g travelling at 500 m/s passes through a wooden block and emerges at 100 m/s. Find energy lost in the block.

$\Delta KE = \frac{1}{2}(0.02)(500^2 - 100^2) = \frac{1}{2}(0.02)(250000 - 10000) = \frac{1}{2}(0.02)(240000) = 2400$ J.

Q7. An engine of power 10 kW pulls a train at 36 km/h on a level track. Find the friction force.

$v = 36 \text{ km/h} = 10$ m/s. $P = Fv$ , so $F = P/v = 10000/10 = 1000$ N.

Q8. A spring of constant 500 N/m is compressed 10 cm. A 0.5 kg ball is placed against it and released. Find the ball’s speed when it leaves the spring.

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$ . $v = x\sqrt{k/m} = 0.1\sqrt{500/0.5} = 0.1\sqrt{1000} = 0.1 \times 31.6 = 3.16$ m/s.

Exam Weightage

Exam	Typical weight	Key topics
CBSE Class 11	6–8 marks	Work-energy theorem, conservation, power
JEE Main	1–2 questions	Collisions, variable force, power
JEE Advanced	1 question	Complex energy conservation, chain/rope problems

Why energy methods beat Newton’s laws

Energy methods shine when:

Forces vary with position (springs, gravity at large distances)
You care about speeds, not accelerations
The path is complex but endpoints are known (roller coasters)
Collisions occur (momentum + energy give two equations for two unknowns)

Newton’s laws are better when you need forces, accelerations, or time.

Potential energy curves

A potential energy curve $U(x)$ encodes all the physics of a conservative system:

Equilibrium points are where $dU/dx = 0$ (force = 0)
Stable equilibrium: $U$ is a minimum ( $d^2U/dx^2 > 0$ ) — particle oscillates
Unstable equilibrium: $U$ is a maximum ( $d^2U/dx^2 < 0$ ) — particle moves away
Total energy line: $E = KE + U$ , so $KE = E - U(x)$ . The particle can only exist where $E \geq U(x)$ .

JEE Advanced often gives a potential energy curve and asks: “At which points is the particle in equilibrium? Which are stable?” Look for minima (stable) and maxima (unstable). The turning points are where $U(x) = E$ .