Work, Energy and Power — NEET Notes

The notion of work in physics

In everyday language, "work" covers everything from carrying bricks to memorising biology. In physics the word has a precise meaning, and getting it precise is the entire point of this chapter. A force does work on a body only when its point of application is displaced. If you push hard against a rigid brick wall and the wall does not move, your muscles are alternately contracting and relaxing — internal energy is being burned, you get tired — but the work done by you on the wall is zero. Effort is not work.

Let a constant force F act on an object that undergoes a displacement d. The work done by the force is the product of the magnitude of the displacement and the component of the force along that displacement:

W = (F cos θ) d = F · d

Work as the scalar product of force and displacement

Three situations produce zero work — and NEET loves to test the second and third. No work is done if (i) the displacement is zero — a weightlifter holding 150 kg motionless on his shoulders does no work on the load no matter how long he stands there; (ii) the force is zero — a block sliding on a smooth horizontal surface experiences no horizontal force, yet it can travel any distance; (iii) the force and the displacement are perpendicular — the gravitational force on a block sliding horizontally does no work; the centripetal force on the Moon (treating its orbit as circular) does no work because the displacement is tangential while the force is radial.

Work can also be negative. Whenever the angle between force and displacement lies between 90° and 180°, cos θ is negative. Friction is the canonical example: kinetic friction always opposes the relative motion, so the angle is 180° and the work it does on the moving body is always negative. The SI unit of work is the joule (J = N m), and the dimensional formula is [M L² T⁻²] — the same as every other form of energy.

Kinetic energy and the work-energy theorem

The work-energy theorem is the bridge between dynamics and energy methods. Start with the one-dimensional kinematic relation under constant acceleration, v² − u² = 2as. Multiply both sides by m/2 and use Newton's second law F = ma:

½ m v² − ½ m u² = F · s

Newton's second law, integrated over displacement

The left side is the change in the quantity ½mv², which we call the kinetic energy K. The right side is the work W done by the net force. Generalised to three dimensions and to varying forces, this becomes the work-energy (WE) theorem.

K_f − K_i = W_net

The work-energy theorem — change in kinetic energy equals work done by the net force

Kinetic energy is a scalar quantity defined as K = ½ m v² = (m/2) v·v. It is always non-negative and depends only on the magnitude of velocity, not on its direction. The theorem holds in any inertial frame, for any kind of force, and over any displacement — though the WE theorem is an integrated form of Newton's second law and so it carries less temporal information than the second law itself.

The power of the theorem lies in what it lets you ignore. To find the speed of a body at the end of a process, you do not need to know the intermediate velocities, the time taken, or the geometry of the path — only the total work done on it. NCERT's raindrop example makes this concrete: a 1 g drop falls 1 km and lands at 50 m/s. The gravitational work is mgh = 10 J. The change in kinetic energy is ½(10⁻³)(50)² = 1.25 J. By the WE theorem, the work done by the (otherwise unknown) air resistance must be 1.25 − 10 = −8.75 J. We have computed the work done by a force whose detailed law we never wrote down.

Work done by a constant force

For a constant force the integral is trivial: W = F · d = F d cos θ. The sign of the work tells you whether energy is being added to or removed from the body's kinetic energy. NCERT's cyclist example: a cyclist comes to a skidding stop in 10 m against a 200 N stopping force directed opposite to the motion. Work done by the road on the cycle is 200 × 10 × cos 180° = −2000 J, exactly the kinetic energy that has been removed. The cycle does not do +2000 J of work on the road, however: by Newton's third law the force on the road equals 200 N in magnitude, but the road undergoes no displacement, so the work done on the road by the cycle is zero. This asymmetry — equal forces, unequal work — is one of NCERT's "points to ponder".

Work done by a variable force

A constant force is the exception, not the rule. The force that a spring exerts depends on how much it has been stretched. The force on a falling raindrop depends on its speed. The retarding force on a body in a rough patch may vary with position. For these cases, divide the displacement into infinitesimal steps Δx so small that F can be treated as constant over each, sum the work done in each step, and take the limit. The sum becomes a definite integral.

W = ∫_xi^x_f F(x) dx

Work done by a one-dimensional variable force — the area under the F–x curve

The work-energy theorem extends seamlessly to variable forces. Differentiating K = ½mv² with respect to time gives dK/dt = mv(dv/dt) = vF = F(dx/dt). So dK = F dx, and integrating from x_i to x_f recovers K_f − K_i = ∫F(x)dx = W. The theorem holds for any force law, however complicated.

The concept of potential energy

The word "potential" suggests stored capacity. A stretched bowstring possesses potential energy; when released, the arrow flies off at great speed. The earth's crust, scarred with fault lines, behaves like a vast network of compressed springs — potential energy that earthquakes release. Potential energy is the energy a body holds by virtue of its position or configuration, ready to be converted to kinetic energy when constraints are removed.

Mathematically, potential energy V(x) can be defined only for a special class of forces — conservative forces — whose work depends only on the initial and final positions, not on the path taken. For such a force in one dimension, V(x) is defined so that

F(x) = −dV/dx   ⇔   ΔV = − ∫ F(x) dx

Potential energy as the negative integral of a conservative force

The minus sign matters. The work done by the gravitational force in lowering a body of mass m by a height h is +mgh, so the potential energy decreases by mgh. Lifting the body to height h above the chosen reference level stores +mgh of gravitational potential energy. Near the earth's surface, where we treat g as constant, V_grav(h) = mgh.

Three equivalent definitions of a conservative force, as NCERT lists them: (i) it can be derived from a scalar potential V(x) by F = −dV/dx; (ii) the work it does between any two points is independent of the path; (iii) the work it does over any closed path is zero. Gravity, the spring force, and electrostatic forces are conservative. Friction is not — it dissipates energy as heat, and no potential energy function can be associated with it.

Conservation of mechanical energy

If the only forces acting on a system are conservative, the total mechanical energy E = K + V is conserved at every point of the motion. This follows directly from the work-energy theorem. For a body under a conservative force, the WE theorem gives ΔK = F(x)Δx. But for a conservative force, F(x)Δx = −ΔV. So ΔK + ΔV = 0, which means Δ(K + V) = 0 — the sum is a constant.

K_i + V(x_i)  =  K_f + V(x_f)  =  constant

Conservation of mechanical energy under conservative forces

The classic illustration is a ball of mass m dropped from a cliff of height H. At the top, the energy is purely potential: E_H = mgH. At an intermediate height h, it has both forms: E_h = mgh + ½mv_h². At the ground, the energy is purely kinetic: E₀ = ½mv_f². Setting E_H = E₀ gives v_f = √(2gH), and setting E_H = E_h recovers v_h = √(2g(H − h)) — familiar kinematic results obtained without writing a single force equation.

Three categories of mechanical energy carry NEET marks every year. They obey one universal book-keeping rule, and the only thing that changes from problem to problem is which of them is non-zero at the start and the end.

When friction is present, the energy ledger has to absorb the loss. NCERT's car-on-spring problem makes this explicit: a 1000 kg car at 5 m/s hits a spring of constant k = 5.25 × 10³ N/m. On a smooth surface, the maximum compression x_m follows from ½mv² = ½kx_m², giving x_m = 2.00 m. With a coefficient of friction μ = 0.5, the change in kinetic energy now equals the work done by the spring and friction together, ½mv² = ½kx_m² + μmgx_m, which gives x_m = 1.35 m — less than the frictionless answer, exactly as one expects.

Potential energy of a spring

The spring force is the textbook example of a position-dependent conservative force. For an ideal spring with one end fixed and a block attached to the other, the spring exerts a force on the block proportional to the block's displacement x from the natural length, directed back towards equilibrium:

F_s = −kx   (Hooke's law)

k is the spring constant, in N m⁻¹. Stiff springs have large k; soft springs have small k.

Because the force varies linearly with x, the work done by the spring as the block moves from 0 to x_m is the area of a triangle on the F-x graph: W_s = −½kx_m². Using the integral form, W_s = ∫₀^x_m (−kx) dx = −½kx_m². The potential energy stored in the spring is the negative of this:

The spring force has every property of a conservative force. The work it does in a complete cycle (stretched then released back to the natural length) is zero. The work from any x_i to any x_f depends only on the endpoints: W_s = ½kx_i² − ½kx_f². And one can verify that −dV/dx = −kx = F_s, which closes the loop.

A block of mass m pulled to extension x_m and released will oscillate between +x_m and −x_m, swapping potential and kinetic energy at every point but keeping their sum constant. The maximum speed, attained at x = 0, follows from ½mv_m² = ½kx_m², giving v_m = x_m√(k/m). This is the basis of simple harmonic motion, which you will meet again later in oscillations.

Power

Power is the rate at which work is done. A person who climbs four floors is fit; one who climbs them quickly is more fit. The average power over a time interval t is P_av = W/t. The instantaneous power is the limiting value as the interval shrinks:

P = dW/dt = F · v

Instantaneous power as the dot product of force and velocity

The SI unit is the watt (W = 1 J s⁻¹), named after James Watt. One horsepower equals 746 W. Bills for electricity use the kilowatt-hour, which is a unit of energy, not power: 1 kWh = 1 kW × 1 hour = 10³ × 3600 J = 3.6 × 10⁶ J. A 100 W bulb burning for 10 hours consumes exactly 1 kWh.

NEET 2022's lift problem is a clean application: a 2000 kg lift moving up at 1.5 m/s against 3000 N of friction. The motor must lift the weight (2000 × 10 = 20000 N) and overcome friction (3000 N), so it pushes upward with 23000 N at 1.5 m/s. The minimum power is P = Fv = 23000 × 1.5 = 34500 W. Power problems are nearly always F·v in disguise — recognising the dot-product form short-circuits a lot of algebra.

Collisions in one and two dimensions

A collision is any process in which two bodies exert large mutual forces on each other over a short time. Billiard balls hitting each other, neutrons scattering off atomic nuclei, alpha particles deflecting near gold atoms — all are collisions in the physicist's sense, whether or not the bodies actually touch. Two laws govern every collision; one of them sometimes fails.

Momentum is conserved in every collision. The argument is short and exact: by Newton's third law, the mutual impulsive forces during the collision are equal and opposite, so the impulses Δp₁ and Δp₂ they impart are equal and opposite, hence Δp₁ + Δp₂ = 0. This is true even though the forces themselves vary wildly during the brief contact time Δt. Kinetic energy is conserved only in elastic collisions. Whenever deformation, heat, or sound are produced, kinetic energy is lost.

One-dimensional collisions

Take a particle of mass m₁ moving with velocity v_1i colliding head-on with a stationary mass m₂. For a perfectly inelastic collision (bodies stick together), momentum conservation gives the joint final velocity v_f = m₁v_1i/(m₁ + m₂). The kinetic energy loss is ΔK = ½ · m₁m₂v_1i²/(m₁+m₂) — always positive, as expected.

For an elastic head-on collision, simultaneously imposing momentum and kinetic energy conservation gives the standard pair of post-collision velocities:

v_1f = (m₁ − m₂)/(m₁ + m₂) · v_1i     v_2f = 2m₁/(m₁ + m₂) · v_1i

1D elastic collision with m₂ initially at rest

Two limiting cases NEET tests repeatedly. Case I — equal masses (m₁ = m₂): v_1f = 0 and v_2f = v_1i. The incoming body stops dead and the target shoots off with the incoming velocity. This is what you see in pool when a moving ball squarely hits a stationary one. Case II — heavy target (m₂ ≫ m₁): v_1f ≈ −v_1i and v_2f ≈ 0. The light projectile bounces back with reversed velocity; the heavy target is essentially undisturbed.

The neutron-moderation example in NCERT is a beautiful application. A fast neutron (≈10⁷ m/s) in a reactor must be slowed to ≈10³ m/s to react with uranium. By colliding elastically with a light moderator nucleus (mass close to the neutron's), the neutron transfers most of its kinetic energy in a single collision. The fractional energy retained by the neutron is f₁ = (m₁−m₂)²/(m₁+m₂)². For deuterium (m₂ = 2m₁), f₁ = 1/9 — about 89% of the energy goes to the deuteron in a single head-on hit. Hence heavy water and graphite are used as reactor moderators.

Two-dimensional (glancing) collisions

If the velocity of the projectile does not pass through the centre of the stationary target, the resulting collision is two-dimensional and the bodies fly off in different directions. Momentum is conserved in both components. Choosing the plane of motion as the x-y plane:

x: m₁v_1i = m₁v_1f cos θ₁ + m₂v_2f cos θ₂
y: 0 = m₁v_1f sin θ₁ − m₂v_2f sin θ₂

Momentum conservation in 2D, with m₂ initially at rest

For an elastic 2D collision of two equal masses with one at rest, momentum and kinetic energy conservation together force the two outgoing velocities to make a right angle: θ₁ + θ₂ = 90°. Every billiards player knows the rule even if they don't know the proof — strike a target ball obliquely and the cue and target separate at 90°. NCERT works this out with the cue going off at 53° when the target enters the pocket at 37°.

Coefficient of restitution

The coefficient of restitution e is the universal index of how "bouncy" a collision is. For a 1D collision along the line of impact:

e = (velocity of separation) / (velocity of approach) = (v_2f − v_1f) / (u_1i − u_2i)

Newton's experimental law of restitution

For a perfectly elastic collision e = 1; for a perfectly inelastic collision (bodies stick) e = 0; for every real collision, 0 < e < 1. NEET 2018 tested this directly: a block of mass m hits a stationary block of mass 4m and comes to rest. Momentum conservation gives v' = v/4 for the 4m block. The velocity of approach is v, the velocity of separation is v/4 − 0 = v/4, so e = (v/4)/v = 0.25.

A useful consequence of the restitution definition: a ball dropped from height h on a flat floor returns to a height e²h after one bounce, e⁴h after two bounces, and so on. The kinetic energy retained after each impact is multiplied by e², so a ball with e = 0.9 loses about 19% of its mechanical energy per bounce.