Work, Energy & Power
Work-energy theorem, conservation of energy, power & collisions
Work and Energy
Work and the Work-Energy TheoremTopic 1
In physics, work has a precise meaning very different from everyday speech. Work is done by a force only when its point of application moves, and only the component of force along the displacement counts: $W = F\,s\cos\theta$, where $\theta$ is the angle between force and displacement. Work is a scalar, measured in joules ($1\ \text{J} = 1\ \text{N}\cdot\text{m}$). Holding a heavy bag still may tire you, but since there is no displacement, the work done on the bag is zero.
The sign of work matters in NEET. Work is positive when the force has a component along the motion (a falling stone, gravity doing positive work), negative when it opposes the motion (friction, or gravity on a rising body), and zero when the force is perpendicular to the displacement. This last case explains why the centripetal force does no work in circular motion and why the normal reaction does no work on a body sliding along a surface — both are perpendicular to the motion.
When the force varies, work is the area under the force-displacement graph. For a spring obeying Hooke's law ($F = kx$), this area gives the work done in stretching or compressing it by $x$ as $W = \tfrac{1}{2}kx^{2}$. Reading such areas is a quick, calculation-free way to compare work in NEET graph questions.
The central result is the work-energy theorem: the net work done on a body equals the change in its kinetic energy, $W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv^{2} - \tfrac{1}{2}mu^{2}$. This single statement links force and motion to energy and often replaces the equations of motion entirely. If the net work is positive the body speeds up; if negative (as with braking friction) it slows down. Because it deals with the net work, the theorem holds whether forces are constant or variable, conservative or not.
| Angle θ | Work done |
|---|---|
| $0^{\circ}$ | maximum, positive ($W = Fs$) |
| $90^{\circ}$ | zero |
| $180^{\circ}$ | negative ($W = -Fs$) |
A force of $10\ \text{N}$ pulls a box $4\ \text{m}$ along the floor at $60^{\circ}$ to the horizontal. Find the work done.
Show solution
$W = Fs\cos\theta = 10\times4\times\cos60^{\circ} = 10\times4\times0.5 = 20\ \text{J}$.
A $2\ \text{kg}$ body speeds up from $3\ \text{m/s}$ to $5\ \text{m/s}$. Find the net work done on it.
Show solution
By the work-energy theorem, $W = \tfrac{1}{2}m(v^{2}-u^{2}) = \tfrac{1}{2}(2)(25-9) = 16\ \text{J}$.
Work done by a force perpendicular to displacement is:
The SI unit of work is the:
The work-energy theorem equates net work to the change in:
Work done in stretching a spring by $x$ (constant $k$) is:
Friction acting opposite to motion does work that is:
NEET tip: Use the work-energy theorem whenever a speed change is involved but you don't need time — it is often faster than the kinematic equations.
Kinetic Energy, Potential Energy and ConservationTopic 2
Energy is the capacity to do work, and mechanics deals chiefly with two mechanical forms. Kinetic energy is energy of motion, $KE = \tfrac{1}{2}mv^{2}$ — note it grows with the square of speed, so doubling speed quadruples kinetic energy, a fact that explains why braking distance rises sharply with speed. A useful link for NEET is $KE = \dfrac{p^{2}}{2m}$, connecting kinetic energy to momentum $p$.
Potential energy is stored energy of position or configuration. Gravitational PE near the Earth's surface is $PE = mgh$, taken relative to a chosen reference level. Elastic PE stored in a stretched or compressed spring is $\tfrac{1}{2}kx^{2}$. Potential energy is defined only for conservative forces (gravity, spring force), for which the work done depends only on the endpoints, not the path taken.
This distinction matters. For a conservative force, the work in a round trip is zero and energy can be stored and recovered. For a non-conservative force like friction, the work depends on the path and energy is dissipated as heat — it cannot be recovered. NEET frequently asks you to classify a force as conservative or not.
The law of conservation of mechanical energy states that if only conservative forces act, the total $KE + PE$ stays constant. A freely falling body steadily converts potential into kinetic energy with their sum unchanged; a pendulum trades the two back and forth. The broader law of conservation of energy says energy is never created or destroyed, only transformed — so when friction is present, the 'lost' mechanical energy reappears as heat, and the total energy is still conserved. Setting (energy at A) = (energy at B), with a heat term when friction acts, solves most NEET energy problems in one line.
| Form of energy | Expression |
|---|---|
| Kinetic energy | $\tfrac{1}{2}mv^{2} = p^{2}/2m$ |
| Gravitational PE | $mgh$ |
| Elastic (spring) PE | $\tfrac{1}{2}kx^{2}$ |
A body of mass $1\ \text{kg}$ is dropped from a height of $5\ \text{m}$. Find its speed just before hitting the ground. ($g = 10\ \text{m/s}^2$)
Show solution
By conservation of energy, $mgh = \tfrac{1}{2}mv^{2}$, so $v = \sqrt{2gh} = \sqrt{2\times10\times5} = 10\ \text{m/s}$.
If the momentum of a body is doubled, by what factor does its kinetic energy change?
Show solution
Since $KE = \dfrac{p^{2}}{2m}$, doubling $p$ multiplies $KE$ by $2^{2} = 4$. The kinetic energy becomes four times larger.
If speed is doubled, kinetic energy becomes:
Potential energy can be defined only for:
For a conservative force, the work done over a closed path is:
A pendulum's energy at the lowest point is mostly:
Mechanical energy is conserved when:
NEET tip: When friction is present, mechanical energy is NOT conserved — write (initial energy) = (final energy) + (heat produced = friction × distance).
Power and Collisions
PowerTopic 3
Power is the rate of doing work, or equivalently the rate of energy transfer: $P = \dfrac{W}{t}$. Its SI unit is the watt ($1\ \text{W} = 1\ \text{J/s}$). Two machines may do the same total work, but the one that does it in less time is more powerful. A common practical unit is the kilowatt ($10^{3}$ W) and the horsepower ($1\ \text{hp} \approx 746\ \text{W}$); the commercial unit of energy, the kilowatt-hour, is power × time and equals $3.6\times10^{6}$ J.
Average power is total work divided by total time, while instantaneous power is the rate at a particular moment, which can be written neatly as $P = \vec{F}\cdot\vec{v} = Fv\cos\theta$. This form is very useful: it shows that for a vehicle moving at constant speed against a resistive force, the engine power equals force times speed.
The dot-product form explains everyday observations NEET likes to probe. A car climbing a hill at steady speed must deliver more power than on level ground because the engine force now also balances the component of gravity along the slope. When a vehicle of fixed engine power accelerates, its driving force $F = P/v$ decreases as speed rises — which is why acceleration tapers off and a maximum speed is eventually reached, where the driving force just balances resistance.
Dimensionally, power has the formula $[\text{ML}^{2}\text{T}^{-3}]$. A frequent NEET task is to compute the power of a pump or motor: lifting water of mass $m$ through height $h$ in time $t$ requires power $P = \dfrac{mgh}{t}$, and if the water also leaves with speed $v$, an extra $\dfrac{1}{2}mv^{2}/t$ of kinetic-energy power is needed. Keeping track of which energy is being supplied per second is the key skill.
| Quantity | Relation |
|---|---|
| Average power | $P = W/t$ |
| Instantaneous power | $P = \vec{F}\cdot\vec{v} = Fv\cos\theta$ |
| 1 kWh | $3.6\times10^{6}\ \text{J}$ |
A pump lifts $600\ \text{kg}$ of water to a height of $10\ \text{m}$ in $20\ \text{s}$. Find its power. ($g = 10\ \text{m/s}^2$)
Show solution
$P = \dfrac{mgh}{t} = \dfrac{600\times10\times10}{20} = \dfrac{60000}{20} = 3000\ \text{W} = 3\ \text{kW}$.
A car of engine power $50\ \text{kW}$ moves at constant $25\ \text{m/s}$. Find the resistive force it overcomes.
Show solution
At constant speed, $P = Fv$, so $F = \dfrac{P}{v} = \dfrac{50000}{25} = 2000\ \text{N}$.
The SI unit of power is the:
Instantaneous power can be written as:
1 kilowatt-hour equals:
For a fixed-power engine, the driving force as speed increases:
The dimensional formula of power is:
NEET tip: For constant-speed motion, engine power = resistive force × speed. For lifting, use $P = mgh/t$ and add any kinetic-energy rate if the load also gains speed.
CollisionsTopic 4
A collision is a brief, strong interaction between bodies. In every collision, provided no external force acts, linear momentum is conserved — this is the master equation, true regardless of the collision type, because the forces during impact are internal. What distinguishes collisions is whether kinetic energy is also conserved.
In a perfectly elastic collision, both momentum and kinetic energy are conserved. Colliding gas molecules and ideal billiard balls approximate this. A striking special case is a one-dimensional elastic collision between equal masses where one is initially at rest: the moving body stops and the stationary one moves off with the full original velocity — a complete exchange of velocities, often tested in NEET. A very light body bouncing off a very heavy one reverses its velocity almost unchanged.
In a perfectly inelastic collision, the bodies stick together and move with a common velocity; momentum is conserved but kinetic energy is maximally lost (converted to heat, sound and deformation). A bullet embedding in a block is the classic example. Most real collisions are partially inelastic — somewhere between the two extremes — with some kinetic energy lost.
The degree of elasticity is measured by the coefficient of restitution $e = \dfrac{\text{relative speed of separation}}{\text{relative speed of approach}}$. It ranges from $e = 1$ for a perfectly elastic collision to $e = 0$ for a perfectly inelastic one. A familiar application is a ball dropped from height $h$ that rebounds to $h' = e^{2}h$, so successive bounce heights fall geometrically. Using momentum conservation together with the value of $e$ resolves almost any NEET collision numerical.
| Collision | Momentum | Kinetic energy | $e$ |
|---|---|---|---|
| Elastic | conserved | conserved | 1 |
| Inelastic | conserved | not conserved | $0 < e < 1$ |
| Perfectly inelastic | conserved | maximally lost | 0 |
A $2\ \text{kg}$ ball at $6\ \text{m/s}$ collides head-on and sticks to a stationary $4\ \text{kg}$ ball. Find their common velocity.
Show solution
Perfectly inelastic, so momentum is conserved: $2\times6 + 4\times0 = (2+4)v$, giving $v = \dfrac{12}{6} = 2\ \text{m/s}$.
A ball dropped from $2\ \text{m}$ rebounds to $0.5\ \text{m}$. Find the coefficient of restitution.
Show solution
$h' = e^{2}h$, so $e = \sqrt{h'/h} = \sqrt{0.5/2} = \sqrt{0.25} = 0.5$.
In all collisions (no external force), the conserved quantity is:
In a perfectly elastic collision, what is also conserved?
For a perfectly inelastic collision, the coefficient of restitution is:
In a 1-D elastic collision of equal masses (one at rest), after impact:
A ball rebounds to $e^2 h$ of its drop height. If $e = 0.8$ and $h = 100\ \text{cm}$, the rebound height is:
NEET tip: Always write momentum conservation first. Add kinetic-energy conservation ONLY for elastic collisions, or use the coefficient of restitution $e$ for the in-between cases.
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