Nearly every year NEET carries one or two questions from this chapter, and it also silently supports rotational motion, gravitation, SHM and even thermodynamics — because "energy stays constant unless something takes it away" is the single most powerful shortcut in mechanics. Master the definitions carefully here and you will solve many later problems without touching a single equation of motion.
What Physics Means by "Work"
In everyday speech, holding a heavy suitcase for ten minutes is hard work. In physics it is zero work. Work is done only when a force succeeds in producing displacement along its own line of action.
For a constant force $\vec{F}$ producing a displacement $\vec{d}$:
$$W = \vec{F}\cdot\vec{d} = Fd\cos\theta$$
where $\theta$ is the angle between the force and the displacement. Work is a scalar even though it is built from two vectors, and its SI unit is the joule (1 J = 1 N m). Its dimensional formula is $[ML^2T^{-2}]$.
The $\cos\theta$ factor decides the sign, and the sign carries physical meaning:
- $\theta < 90^\circ$ → positive work: the force is helping the motion and feeding energy into the body (gravity on a falling stone).
- $\theta = 90^\circ$ → zero work: the force is perpendicular to motion (centripetal force in uniform circular motion; normal reaction on a horizontal floor; magnetic force on a moving charge).
- $\theta > 90^\circ$ → negative work: the force opposes motion and drains energy (friction, or gravity on a body being lifted).
Zero work also arises in two other ways: when the force itself is zero, or when there is no displacement at all (pushing a wall). Note that displacement, not distance, appears in the formula — so for a body carried around a closed loop, gravity does zero net work no matter how complicated the path.
Work done by a variable force. If the force changes with position, divide the path into slices so small that $\vec{F}$ is effectively constant over each, add the contributions, and take the limit:
$$W = \int_{x_1}^{x_2} F_x , dx$$
Graphically, work is the area under the force–displacement graph (area below the axis counted as negative). This graphical route is a favourite in exams because it lets you handle forces given as diagrams rather than formulas. For a spring obeying Hooke's law, $F = -kx$, and stretching it from 0 to $x$ requires work $\tfrac12 kx^2$ against the spring, so the spring itself does $-\tfrac12 kx^2$.
Kinetic Energy and the Work–Energy Theorem
A moving body can do work by virtue of its motion; this capacity is its kinetic energy,
$$K = \tfrac12 mv^2 = \frac{p^2}{2m}$$
Kinetic energy is always positive (or zero) and is a scalar. The second form, in terms of momentum $p = mv$, is extremely handy: if two bodies have equal momenta, the lighter one has more KE; if they have equal KE, the heavier one has more momentum.
The work–energy theorem states that the net work done by all forces acting on a particle equals the change in its kinetic energy:
$$W_{\text{net}} = K_f - K_i = \Delta K$$
A quick derivation for one dimension with constant acceleration: from $v^2 - u^2 = 2as$, multiply through by $m/2$ to get $\tfrac12 mv^2 - \tfrac12 mu^2 = (ma)s = Fs$. The theorem is far more general than this derivation suggests — it holds for variable forces and curved paths too.
Two features make it a problem-solving weapon:
- Time never appears. If a question gives you speeds and distances but not time, reach for this theorem.
- It uses the net work, so you may either add up work by each force separately or compute the resultant force first.
Typical uses: finding the stopping distance of a car (the retarding work $-fs$ equals $-\tfrac12 mv^2$, so $s \propto v^2$ — doubling the speed quadruples the stopping distance); finding the depth a bullet penetrates into a block; finding the speed at the bottom of a curved frictionless slide.
Potential Energy and Conservative Forces
Potential energy is energy stored because of position or configuration. It is only definable for a special class of forces called conservative forces, for which the work done depends solely on the initial and final positions, not on the path taken (equivalently, the work around any closed path is zero). Gravity, spring force, and electrostatic force are conservative; friction, air drag and viscous forces are non-conservative (dissipative) — drag a block in a circle back to its start and friction has robbed it of energy in proportion to the total path length.
For a conservative force, potential energy is defined by
$$\Delta U = -W_{\text{conservative}}, \qquad F_x = -\frac{dU}{dx}$$
The minus sign says the force pushes a body "downhill" on the potential energy graph. Only changes in $U$ matter physically; the zero level is our free choice.
Standard forms to memorise:
- Gravitational PE near the Earth's surface: $U = mgh$ (valid only where $g$ is effectively constant).
- Elastic PE of a spring: $U = \tfrac12 kx^2$, where $x$ is the deformation from the natural length (compression and extension store equal energy).
Conservation of mechanical energy. If only conservative forces do work, then $\Delta K + \Delta U = 0$, so
$$K + U = \text{constant}$$
For a body sliding down a smooth incline or track of height $h$, $v = \sqrt{2gh}$ regardless of the shape of the track — the answer depends on the vertical drop alone. For a vertically thrown ball, KE at the bottom converts fully to PE at the top.
When friction or other dissipative forces are present, mechanical energy is not conserved, but the more general law survives:
$$W_{\text{non-conservative}} = \Delta K + \Delta U$$
Nothing is truly "lost" — the missing mechanical energy appears as heat, sound or deformation. This is the content of the law of conservation of energy: energy can only be transformed, never created or destroyed. Einstein's relation $E = mc^2$ extends the bookkeeping to mass itself, which matters in nuclear reactions.
Reading potential energy curves is a skill NEET tests. On a $U$ vs $x$ graph: points where the slope is zero are equilibrium points; a minimum of $U$ is stable equilibrium, a maximum is unstable. Since $K = E - U$ must be non-negative, regions where $U > E$ are forbidden, and the points where $U = E$ are the turning points of the motion.
Power
Power measures how fast work is done — the rate of energy transfer.
$$P_{\text{avg}} = \frac{W}{t}, \qquad P_{\text{inst}} = \frac{dW}{dt} = \vec{F}\cdot\vec{v} = Fv\cos\theta$$
The SI unit is the watt (1 W = 1 J s⁻¹); dimensions $[ML^2T^{-3}]$. Practical units: 1 horsepower = 746 W. Note that the kilowatt-hour is a unit of energy, not power: 1 kWh = 3.6 × 10⁶ J.
The relation $P = Fv$ explains everyday driving: a vehicle engine delivering constant power must supply less force at high speed, which is why drivers shift to a lower gear (more force, less speed) to climb a slope. For a machine lifting water or sand at a steady rate, useful power = (mass lifted per second) × $g$ × height, and for a pump pushing out water at speed $v$, the kinetic power delivered is $\tfrac12 (dm/dt) v^2$.
Efficiency is the ratio of useful output energy (or power) to input energy (or power), usually expressed as a percentage.
Collisions
A collision is a brief, strong interaction in which the mutual forces are so large that external forces can be neglected during contact. Therefore:
Linear momentum is conserved in every collision. Kinetic energy may or may not be.
Classification:
- Perfectly elastic: both momentum and kinetic energy conserved (collisions of hard steel balls, atomic and subatomic collisions are close to this).
- Inelastic: momentum conserved, KE decreases; some energy goes into heat, sound or permanent deformation.
- Perfectly inelastic: the bodies stick together and move with a common velocity; the KE loss is the maximum allowed by momentum conservation.
The coefficient of restitution quantifies this: $e = \dfrac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$, with $e = 1$ for perfectly elastic, $e = 0$ for perfectly inelastic, and $0 < e < 1$ in between. For a ball dropped from height $h$ rebounding to $h'$, $e = \sqrt{h'/h}$.
One-dimensional elastic collision. For masses $m_1, m_2$ with initial velocities $u_1, u_2$:
$$v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 + \frac{2m_2}{m_1+m_2}u_2, \qquad v_2 = \frac{m_2-m_1}{m_1+m_2}u_2 + \frac{2m_1}{m_1+m_2}u_1$$
Special cases worth remembering (take $u_2 = 0$):
- Equal masses: the velocities are simply exchanged — the incoming ball stops dead and the target moves off with $u_1$.
- Very heavy projectile hitting a light target ($m_1 \gg m_2$): the heavy body sails on almost unchanged, the light one flies off at nearly $2u_1$.
- Light projectile hitting a very heavy target ($m_1 \ll m_2$): the projectile rebounds with almost the same speed reversed, the heavy target barely moves (a ball bouncing off a wall).
A useful shortcut for 1-D elastic collisions: relative velocity of approach = relative velocity of separation, i.e. $u_1 - u_2 = v_2 - v_1$. Combined with momentum conservation, this pair of linear equations is faster than the quadratic energy equation.
For a perfectly inelastic head-on collision with $u_2 = 0$, the common velocity is $v = \dfrac{m_1u_1}{m_1+m_2}$ and the fractional KE lost is $\dfrac{m_2}{m_1+m_2}$ — note that a lighter target means less energy loss.
In two dimensions, momentum is conserved separately along each axis; for an elastic collision between equal masses when one is initially at rest, the two bodies always separate at 90° to each other.
Common Mistakes and Exam Traps
- Confusing distance with displacement in $W = Fd\cos\theta$. Only the displacement component along the force counts for conservative forces; but for friction, the work done is $-f \times$ (total path length), not $-f \times$ displacement. That asymmetry is exactly what makes friction non-conservative.
- Assuming energy conservation when friction or a "sticking" event is present. In perfectly inelastic collisions and in any problem mentioning rough surfaces, momentum (or the work–energy theorem) is the safe tool; mechanical energy conservation is wrong.
- Treating power as constant when it isn't, or mixing units. For a body falling freely, the instantaneous power of gravity $mgv$ grows with time even though the force is constant. Also watch out for unit mismatches: if speed is given in km/h and force in newtons, convert to SI before computing power in watts — mixing units here is a very common arithmetic trap, especially in problems dressed up as "a pump lifts water" or "a motor pulls a load up an incline."
- Assuming the simple work–energy theorem applies unchanged to variable-mass or non-inertial situations. $W_{net} = \Delta K$ is derived for a fixed mass observed from an inertial frame; problems involving a leaking sand bag, a rocket ejecting fuel, or motion described from inside an accelerating lift require either a modified variable-mass treatment or the explicit inclusion of a pseudo-force before the theorem can be applied safely.