Unit 4: Work, Energy and Power

This unit yields 2–3 MCQs in NEET each year. The questions cluster around three skills: (i) computing work done by various forces (including variable forces and on a spring), (ii) applying conservation of mechanical energy in problems with gravity and springs, and (iii) collision numericals — especially using the coefficient of restitution.

The unit links directly to gravitation, oscillations, and electrostatics, where energy methods reappear.

Concept Map

• Work by a constant or variable force
• Kinetic energy and the work-energy theorem
• Conservative forces and potential energy
  • Gravitational PE
  • Spring PE
• Conservation of mechanical energy
• Power (average and instantaneous)
• Collisions
  • 1D elastic and inelastic
  • 2D elastic (basics)
  • Coefficient of restitution
• Non-conservative forces (friction, drag)

Topic 1: Work

Sub-topic A: Work by a Constant Force

where $\theta$ is the angle between $\vec F$ and the displacement $\vec d$. Work is a scalar with SI unit joule (J $=$ N·m).

Special cases:

• $\theta = 0°$: $W = Fd$ (max positive).
• $\theta = 90°$: $W = 0$ (centripetal force, normal force, magnetic force do no work).
• $\theta = 180°$: $W = -Fd$ (friction opposing motion).

Sub-topic B: Work by a Variable Force

For 3D: $W = \int \vec F \cdot d\vec r$.

Sub-topic C: Work Done by Spring

For a spring of stiffness $k$ stretched/compressed from $x_1$ to $x_2$:

The work done on the spring by the external agent is $+\tfrac{1}{2} k x^2$ when stretched from natural length by $x$.

Sub-topic D: Work Done by Gravity

For a body lifted from height $h_1$ to $h_2$ (near earth surface):

Positive when the body falls, negative when it rises.

Topic 2: Kinetic Energy and Work-Energy Theorem

Sub-topic A: Kinetic Energy

Always positive (or zero). Scalar with units of energy.

Sub-topic B: Work-Energy Theorem

The net work done by all forces equals the change in KE:

This holds for variable forces, in 1D or 3D. NEET frequently asks: a bullet penetrates depth $d_1$ in one block; find depth in a second block — uses $W = -F d$ and $\Delta K$ equality.

Topic 3: Potential Energy and Conservative Forces

Sub-topic A: Conservative Forces

A force is conservative if the work it does around any closed loop is zero, equivalently if the work depends only on initial and final positions. Examples: gravity, spring, electrostatic. Non-conservative: friction, air drag, viscous drag.

For a conservative force $\vec F$ there exists a potential energy $U$ with

Sub-topic B: Gravitational PE (near surface)

with $h$ measured from a chosen reference level. Only differences in $U$ matter physically.

Sub-topic C: Spring PE

Sub-topic D: PE-Force Relation in 1D

If $U(x) = a x^2 - b x$, then $F = -dU/dx = -2 a x + b$. Equilibrium at $F = 0$, i.e. $x = b/(2a)$. Stable equilibrium where $d^2 U/dx^2 > 0$.

Topic 4: Conservation of Mechanical Energy

If only conservative forces act:

NEET-style application: a body of mass $m$ slides from height $h$ down a smooth ramp; speed at bottom $v = \sqrt{2gh}$. With friction $\mu$ over length $L$, energy theorem gives $\tfrac{1}{2} m v^2 = m g h - \mu m g \cos\theta \cdot L$.

For a spring of stiffness $k$ compressed by $x$ and released, the block of mass $m$ leaves the spring with $v = x\sqrt{k/m}$.

Topic 5: Power

Sub-topic A: Definitions

• Average power: $\bar P = W/\Delta t$.
• Instantaneous power: $P = dW/dt = \vec F \cdot \vec v$.

Sub-topic B: Units and Conversions

• SI unit: watt (W) $=$ J/s.
• Horsepower: $1\ \text{hp} = 746\ \text{W}$.
• Kilowatt-hour (kWh) is an energy unit: $1\ \text{kWh} = 3.6 \times 10^6\ \text{J}$.

Sub-topic C: Common Applications

• Car of mass $m$ moving up an incline at constant $v$: $P = m g v \sin\theta$.
• Lift carrying $m$ at constant $v$: $P = m g v$.
• Pump raising water of density $\rho$ at rate $Q$ (m³/s) to height $h$: $P = \rho Q g h$.

Topic 6: Collisions

Sub-topic A: Conservation Laws

In all collisions (elastic, inelastic, perfectly inelastic), linear momentum is conserved (no external impulsive force). KE is conserved only in elastic collisions.

Sub-topic B: Coefficient of Restitution

• $e = 1$: perfectly elastic.
• $0 < e < 1$: inelastic.
• $e = 0$: perfectly inelastic (objects stick).

Sub-topic C: 1D Elastic Collision

For masses $m_1, m_2$ with initial velocities $u_1, u_2$:

Special cases:

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

Sub-topic D: Perfectly Inelastic Collision

Bodies stick together. Final common velocity:

Energy loss:

Sub-topic E: Ball Bouncing on Ground

Ball dropped from height $h_0$. After the $n$-th bounce with coefficient $e$:

Total distance travelled before coming to rest:

Total time:

Sub-topic F: 2D Elastic Collision

When two equal masses collide elastically, with one at rest, they scatter at 90° to each other.

NEET Pattern MCQ Tips

• Work by variable force: integrate or use F-x area.
• Spring problems: $U = \tfrac{1}{2} kx^2$, energy conservation.
• Vertical circle / loop the loop: energy + circular motion combined.
• Collisions in 1D: identify elastic/inelastic and use the velocity-exchange shortcut.
• Bouncing ball: total distance formula appears every few years.
• Assertion-Reason: KE conserved in elastic collision; momentum conserved in all.
• Power: lift, pump, car-on-incline.

Common Confusions and Traps

• Centripetal force does no work because it is perpendicular to velocity.
• Friction can do positive work — e.g., the friction from the ground on a walking person's feet, in the direction of motion.
• KE is not conserved in inelastic collisions, but momentum always is (in absence of external impulse).
• The work done by spring is $-\tfrac{1}{2} k x^2$; the work done on spring is $+\tfrac{1}{2} k x^2$.
• For elastic collisions with equal masses at rest, velocities are exchanged — students often forget this shortcut.
• Power $P = Fv$ holds only when $\vec F$ and $\vec v$ are parallel; in general $P = \vec F \cdot \vec v$.
• A perfectly inelastic collision does not mean all KE is lost — only the maximum compatible with momentum conservation is.

Quick Revision Card

• $W = \vec F \cdot \vec d$; $W = \int F\,dx$ for variable force.
• $W_\text{spring on agent} = +\tfrac{1}{2} k x^2$ (work done on spring); $W_\text{spring on block} = -\tfrac{1}{2} k x^2$.
• Work-energy theorem: $W_\text{net} = \Delta K$.
• $F = -dU/dx$; stable equilibrium where $U$ has a minimum.
• Power: $P = \vec F \cdot \vec v$; $P = m g v$ for lifting at constant speed.
• $e =$ (separation)/(approach).
• Elastic collision, equal masses: velocities exchange.
• Perfectly inelastic: $v = (m_1 u_1 + m_2 u_2)/(m_1 + m_2)$.
• Bouncing ball: $h_n = e^{2n} h_0$; total distance $= h_0(1 + e^2)/(1 - e^2)$.

Worked NEET Examples

Example 1: Block Compressing a Spring

A block of mass 2 kg moving at 4 m/s on a smooth horizontal surface compresses a spring of stiffness $k = 100$ N/m. Maximum compression?

Energy conservation: $\tfrac{1}{2}mv^2 = \tfrac{1}{2}k x_\text{max}^2$.

Example 2: Loop-the-Loop

A ball slides down a smooth track from height $h$ and enters a vertical loop of radius $R$. Minimum $h$ for the ball to complete the loop?

At top of loop, minimum speed: $v_\text{top}^2 = gR$. Energy from start to top:

So $h_\text{min} = 2.5 R$.

Example 3: Pendulum Cut

A pendulum of length $L$ swings in vertical circle, with bob of mass $m$ moving at $v_0$ at the lowest point. The string can sustain tension up to $T_\text{max} = 3mg$. Will the string break?

At the lowest point: $T - mg = m v_0^2/L$, so $T = m(g + v_0^2/L)$. If $v_0^2/L \le 2g$, then $T \le 3mg$ and string survives. Otherwise breaks.

Example 4: Bullet Penetration

A 10 g bullet at 500 m/s penetrates 5 cm into a fixed wooden block. Average resisting force?

Work done against force = initial KE: $F \cdot d = \tfrac{1}{2} mv^2$.

Example 5: Power of an Engine on an Incline

An engine pulls a 1000 kg vehicle up a 30° incline at constant 5 m/s. Frictional force is 100 N. Engine power?

Force needed to maintain motion: $mg\sin\theta + f = 1000 \times 10 \times 0.5 + 100 = 5100$ N.

Power: $P = Fv = 5100 \times 5 = 25500$ W = 25.5 kW.

Derivations Summary

Conservation of Mechanical Energy

For a conservative force, work-energy theorem gives $W = K_f - K_i$. But the work done by a conservative force is $-(U_f - U_i)$. So $K_f + U_f = K_i + U_i$, i.e. total mechanical energy is conserved.

Energy Loss in Perfectly Inelastic Collision

Two masses $m_1, m_2$ approach with $u_1, u_2$, stick after collision moving at $v = (m_1 u_1 + m_2 u_2)/(m_1 + m_2)$.

Initial KE: $\tfrac{1}{2}(m_1 u_1^2 + m_2 u_2^2)$. Final KE: $\tfrac{1}{2}(m_1 + m_2)v^2$. After algebra:

So energy is always lost when bodies stick together.

Maximum Power in a Resistive Circuit

A cell of EMF $\varepsilon$ and internal resistance $r$ connected to external $R$. Current $I = \varepsilon/(R + r)$, power in $R$:

Differentiating with respect to $R$ and setting to zero gives $R = r$ for maximum power, $P_\text{max} = \varepsilon^2/(4r)$, efficiency $= R/(R + r) = 50\%$.

Variable Force / Position-Dependent Forces

If $F$ depends on $x$, work is $W = \int F\,dx$. Example: a particle in an inverse-square field $F = -k/x^2$ moving from $x_1$ to $x_2$. Work done by the field:

This is exactly $-(U_2 - U_1)$ with $U = -k/x$.

Power-Time Profiles

A variable-power engine with $P(t) = P_0 + \alpha t$. Total work done in time $t$:

If applied to a constant-mass body initially at rest, $W = \tfrac{1}{2}mv^2$, giving $v(t)$ implicitly.

For constant power $P$ and constant mass $m$, motion from rest gives $v = \sqrt{2Pt/m}$ and $x = (2P/m)^{1/2}\,(2/3)\,t^{3/2}$. So $x \propto t^{3/2}$ — frequently asked.

Collisions in Two Dimensions

For elastic collision in 2D with $m_1$ hitting $m_2$ (at rest), $m_1 = m_2$: the two scatter at 90° to each other. The total KE and momentum are conserved.

In general, four unknowns ($v_1', v_2'$, two angles) and three equations (two momentum components + KE) — so problem is under-determined unless the geometry of the impact (impact parameter) is specified.

Formula Sheet