Work, Energy and Power

This chapter has a clear build order, and it pays to see the map before the details. It opens with a short mathematical tool, the scalar (dot) product of two vectors, because work is defined using it. It then defines work and kinetic energy for a constant force and proves the work-energy theorem, before generalising both to a force that varies with position using the area under a force-displacement graph and, ultimately, an integral. From there it introduces potential energy as the energy "stored" by a conservative force, works out gravitational and spring potential energy explicitly, and proves the conservation of mechanical energy — the headline result that lets you skip Newton's second law entirely in many problems. Power is treated as a short, self-contained idea: the rate of doing work. The chapter closes with collisions, where the conservation laws of momentum and energy are applied together to work out what happens when two bodies meet, in one dimension and then two.

The scalar product of two vectors

Work needs a way to combine a force vector and a displacement vector into a single number — a scalar — that captures how much of the force acts along the direction of motion. That tool is the scalar product, or dot product, of two vectors $\vec{A}$ and $\vec{B}$:

where: $A$ and $B$ are the magnitudes of the two vectors, and $\theta$ is the angle between them when drawn tail to tail. Because $A$, $B$ and $\cos\theta$ are all scalars, the dot product is a scalar — it has a size but no direction, even though it was built from two vectors that each have direction. Geometrically, $B\cos\theta$ is the projection of $\vec{B}$ onto $\vec{A}$, so $\vec{A}\cdot\vec{B}$ is the magnitude of $\vec{A}$ times the component of $\vec{B}$ that lies along $\vec{A}$ (or equally, the magnitude of $\vec{B}$ times the component of $\vec{A}$ along $\vec{B}$). The dot product is commutative, $\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}$, and distributive, $\vec{A}\cdot(\vec{B}+\vec{C}) = \vec{A}\cdot\vec{B} + \vec{A}\cdot\vec{C}$. In component form, using unit vectors $\hat{i}, \hat{j}, \hat{k}$ that satisfy $\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1$ and $\hat{i}\cdot\hat{j}=\hat{j}\cdot\hat{k}=\hat{k}\cdot\hat{i}=0$, the dot product becomes simply $\vec{A}\cdot\vec{B} = A_xB_x + A_yB_y + A_zB_z$, which is how it is actually computed in problems. Two immediate consequences matter throughout this chapter: $\vec{A}\cdot\vec{A} = A^2$ (a vector dotted with itself gives the square of its magnitude), and $\vec{A}\cdot\vec{B} = 0$ whenever the two vectors are perpendicular — this is exactly why a force perpendicular to the displacement it acts over does zero work.

Work, kinetic energy and the work-energy theorem for a constant force

Consider a constant force $\vec{F}$ acting on an object of mass $m$ that undergoes a displacement $\vec{d}$. The work done by the force is defined as the product of the component of the force along the displacement and the magnitude of the displacement:

where: $\theta$ is the angle between $\vec{F}$ and $\vec{d}$. This single definition explains three "no work" situations that trip up beginners: if the displacement is zero (a weightlifter holding a mass steadily overhead does no work on it, however hard the muscles strain), if the force is zero (a block gliding on a frictionless table has no horizontal force acting, however far it travels), or if force and displacement are mutually perpendicular ($\theta = 90°$, so $\cos\theta = 0$ — this is why gravity does no work on the Moon in an idealised circular orbit, since the gravitational pull is always radial while the velocity is always tangential). Work can be positive (0° ≤ θ < 90°), negative (90° < θ ≤ 180°, as when friction opposes motion), or zero. Since $W = \vec{F}\cdot\vec{d}$ has the same dimensions as energy, $[ML^2T^{-2}]$, work and energy share the SI unit joule (J), named for James Prescott Joule. Kinetic energy measures the capacity of a moving body to do work by virtue of its motion:

where: $m$ is the mass and $v$ the speed; $K$ is a scalar and, unlike work, it can never be negative. Starting from the kinematic relation $v^2 - u^2 = 2as$ for constant acceleration, and generalising to vectors as $v^2 - u^2 = 2\,\vec{a}\cdot\vec{d}$, multiplying both sides by $m/2$ and using Newton's second law $m\vec{a} = \vec{F}$ gives $\tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2 = \vec{F}\cdot\vec{d}$. The left side is the change in kinetic energy and the right side is exactly the work done by the net force, so

This is the work-energy theorem: the change in kinetic energy of a particle equals the work done on it by the net force. It is worth noting what the theorem does not tell you — it says nothing about the time taken, and in two or three dimensions it discards the directional information that Newton's second law carries, because it is a scalar statement. But its power is that it lets you find a final speed without ever solving for acceleration as a function of time, and — as Example 5.2 in the chapter shows for a falling raindrop with an unknown air-resistance force — it can be used even when the exact form of one of the forces is not known, as long as you know the total change in kinetic energy.

Work done by a variable force

A constant force is the exception, not the rule; friction, springs and most real forces change as an object moves. Fig 5.3 in the NCERT text builds the idea by chopping the displacement into small steps $\Delta x$, over each of which the force $F(x)$ is treated as roughly constant, so that $\Delta W = F(x)\,\Delta x$. Summing these rectangular strips of area under the force-versus-displacement curve gives an approximate total work; as the strips are made infinitesimally thin ($\Delta x \to 0$), the sum becomes exact and turns into a definite integral:

where: $F(x)$ is the force as a function of position, and $x_i$, $x_f$ are the initial and final positions. Physically, $W$ is simply the area under the force-displacement graph between $x_i$ and $x_f$, with area below the axis (where $F$ is negative, opposing displacement) counted as negative work — exactly the case for a frictional force plotted against distance.

Adding these two pieces of area, one above the axis and one below, is exactly how you total the work done by two different forces acting over the same displacement, a pattern used repeatedly in problems where an applied force and friction act together.

Potential energy and conservative forces

Some forces have a special property: the work needed to move a body against them does not disappear, it gets "stored" and can be recovered later as kinetic energy — a stretched bowstring, a compressed spring, a raised weight. Such forces are called conservative, and for a conservative force acting in one dimension a potential energy function $V(x)$ can be defined so that

where: the derivative $dV/dx$ is the slope of the potential energy curve, and the minus sign means the force always points from high potential energy toward low potential energy — a ball rolls downhill, not uphill, because the force is minus the slope. Equivalently, $\displaystyle\int_{x_i}^{x_f}F(x)\,dx = V_i - V_f$, so the change in potential energy is the negative of the work done by the force, $\Delta V = -F(x)\Delta x$. A conservative force can be identified in three equivalent ways: its work depends only on the initial and final positions, not on the path taken; its work around any closed path (returning to the start) is zero; or it can be written as minus the derivative of some scalar $V(x)$. Gravity and the ideal spring force both qualify; friction does not, because the work done against friction depends on the length of the path travelled, not just the endpoints, and cannot be recovered. For gravity near Earth's surface, raising a mass $m$ through a height $h$ against the constant force $mg$ stores potential energy $V(h) = mgh$ (taking $V = 0$ at the ground); differentiating confirms $F = -dV/dh = -mg$, the downward gravitational force. For an ideal spring obeying Hooke's law, $F_s = -kx$, where $x$ is the displacement of the block from its equilibrium position and $k$ is the spring constant (stiff springs have large $k$, soft springs small $k$), the same procedure gives the spring's potential energy as $V(x) = \tfrac{1}{2}kx^2$, taking $V=0$ at the natural, unstretched length. In every case the zero of potential energy is a free choice, fixed for convenience at the start of a problem and then held fixed throughout — you cannot change the reference point midway through a calculation.

Conservation of mechanical energy and the spring

Combine the work-energy theorem with the definition of potential energy and one of physics' most useful results appears almost for free: if the only forces doing work on a body are conservative, its total mechanical energy, $E = K + V$, stays constant throughout the motion. The next section derives this formally; physically, it says energy sloshes between kinetic and potential form without loss — a ball dropped from height $H$ is purely potential energy at the top and purely kinetic just before impact, with every intermediate height sharing the total between the two forms, $E_H = mgH = \tfrac{1}{2}mv^2 = E_0$. The block-and-spring system on a smooth horizontal surface shows the same conservation with a different potential energy shape. If the block is pulled to an extension $x_m$ and released from rest, its total mechanical energy at any position $x$ between $-x_m$ and $+x_m$ satisfies $\tfrac{1}{2}kx_m^2 = \tfrac{1}{2}kx^2 + \tfrac{1}{2}mv^2$, so speed and kinetic energy are maximum at the equilibrium position $x = 0$ and fall to zero at the extremes $x = \pm x_m$, where all the energy is potential.

Power

Work tells you how much is done, power tells you how fast. Average power is the ratio of work done to the time taken, $P_{av} = W/t$, and instantaneous power is its limiting, calculus form:

where: $\vec{v}$ is the instantaneous velocity of the point where the force $\vec{F}$ acts. Power, like work and kinetic energy, is a scalar with dimensions $[ML^2T^{-3}]$; the SI unit is the watt (1 W = 1 J/s), named for James Watt, and the older unit horsepower converts as $1\ \text{hp} = 746\ \text{W}$. A useful everyday distinction: the kilowatt-hour (kWh) that appears on an electricity bill is a unit of energy, not power — a 100 W bulb left on for 10 hours consumes $100 \times 10 = 1000\ \text{Wh} = 1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$.

Collisions: elastic and inelastic

A collision is a brief, intense interaction between two bodies. Whatever the details of the impulsive forces during the collision, Newton's third law guarantees that the impulse on one body is equal and opposite to the impulse on the other over the collision time $\Delta t$, so the total linear momentum of the two-body system is conserved: $\Delta \vec{p}_1 + \Delta \vec{p}_2 = 0$, true in every collision without exception. Kinetic energy is a different story. If the bodies rebound without any permanent deformation, sound or heat generated, the total kinetic energy after the collision equals the total before — this is an elastic collision. If the deformation is permanent and the two bodies stick together and move with a common final velocity, it is a completely inelastic collision, and kinetic energy is lost to heat, sound and deformation (though momentum is still conserved). Most real collisions are partly in between: an inelastic collision, where some kinetic energy is lost but the bodies still separate. For a head-on (one-dimensional) elastic collision of mass $m_1$ moving at $v_{1i}$ with a stationary mass $m_2$, both momentum and kinetic energy conservation together give the final speeds worked out in the derivation below. Two special cases are worth remembering by physical intuition alone: if $m_1 = m_2$, the moving mass stops dead and hands its entire velocity to the target ($v_{1f}=0$, $v_{2f}=v_{1i}$); if $m_2 \gg m_1$ (a light ball bouncing off a heavy wall), the heavy mass stays essentially still while the light one simply reverses direction ($v_{2f}\approx 0$, $v_{1f}\approx -v_{1i}$). In two dimensions, momentum conservation along two independent axes plus energy conservation (if elastic) still leaves one angle undetermined by conservation laws alone — it must be measured or given, as in a glancing billiard-ball shot.

Board level - angle between force and displacement using the scalar product

A force $\vec{F} = (3\hat{i}+4\hat{j}-5\hat{k})$ unit acts on a body that undergoes a displacement $\vec{d} = (5\hat{i}+4\hat{j}+3\hat{k})$ unit. Find the angle between the force and the displacement, and the work done by the force. (Following NCERT Example 5.1.)

Step 1. Find the dot product using components: $\vec{F}\cdot\vec{d} = F_xd_x+F_yd_y+F_zd_z = 3(5)+4(4)+(-5)(3) = 15+16-15 = 16$ unit. Since $\vec{F}\cdot\vec{d}=Fd\cos\theta$, this dot product is itself the work done by the force, so $W = 16$ J (taking the units as SI).

Step 2. Find the magnitudes needed for $\cos\theta$: $F^2=\vec{F}\cdot\vec{F}=3^2+4^2+(-5)^2=9+16+25=50$, and $d^2=\vec{d}\cdot\vec{d}=5^2+4^2+3^2=25+16+9=50$. So $F=\sqrt{50}$ and $d=\sqrt{50}$.

Step 3. Solve for the angle: $\cos\theta = \dfrac{\vec{F}\cdot\vec{d}}{Fd} = \dfrac{16}{\sqrt{50}\sqrt{50}} = \dfrac{16}{50} = 0.32$, so $\theta = \cos^{-1}(0.32) \approx 71.3°$. Answer: the work done is 16 J and the angle between force and displacement is $\cos^{-1}(0.32)\approx 71.3°$.

Board level - power delivered by an elevator motor

An elevator, of total mass 1800 kg including passengers, moves upward at a constant speed of 2 m/s. A frictional force of 4000 N opposes its motion. Determine the minimum power the motor must deliver, in watts and in horsepower. (Following NCERT Example 5.10, with $g=10\ \text{m/s}^2$.)

Step 1. Identify every downward force the motor must overcome: the weight $mg = 1800\times10 = 18000$ N, plus the frictional force of 4000 N acting against the upward motion. The total opposing force is $F = 18000+4000 = 22000$ N.

Step 2. Since the elevator moves at constant speed, the motor's upward force must exactly balance this total, so the force the motor supplies is also 22000 N, acting in the same direction as the velocity.

Step 3. Apply $P=\vec{F}\cdot\vec{v}$, and since force and velocity are parallel (both upward) this is simply $P = Fv = 22000\times2 = 44000$ W. Converting using $1\ \text{hp}=746\ \text{W}$: $P = 44000/746 \approx 59$ hp. Answer: the motor must deliver at least 44000 W, or about 59 hp.

Competitive level - maximum spring compression with friction (JEE pattern)

A car of mass 1000 kg moving at 18.0 km/h on a smooth road collides with a horizontally mounted spring of spring constant $k=5.25\times10^3\ \text{N/m}$, and this time the road has friction with coefficient $\mu=0.5$ over the region of contact. Take $g=10\ \text{m/s}^2$. Find the maximum compression of the spring. (Following NCERT Example 5.9.)

Step 1. Convert the speed to SI: $18.0\ \text{km/h} = 5\ \text{m/s}$ (using $36\ \text{km/h}=10\ \text{m/s}$ as a quick benchmark). The change in kinetic energy as the car comes to rest at maximum compression is $\Delta K = 0 - \tfrac{1}{2}mv^2 = -\tfrac{1}{2}(1000)(5)^2 = -12500$ J.

Step 2. With friction present, use the work-energy theorem rather than plain energy conservation, since friction is non-conservative. The net work done on the car as it compresses the spring by $x_m$ is $W = -\tfrac{1}{2}kx_m^2 - \mu mg\,x_m$ (the spring force and friction both oppose the compression, so both contribute negative work).

Step 3. Equate $\Delta K = W$: $-\tfrac{1}{2}mv^2 = -\tfrac{1}{2}kx_m^2 - \mu mg\,x_m$, which rearranges into a quadratic in $x_m$: $kx_m^2 + 2\mu mg\,x_m - mv^2 = 0$. Substituting $\mu mg = 0.5\times1000\times10 = 5000$ N, this becomes $5250\,x_m^2 + 10000\,x_m - 25000 = 0$.

Step 4. Solve the quadratic, keeping only the positive root (compression must be a positive length): $x_m = \dfrac{-10000+\sqrt{10000^2+4(5250)(25000)}}{2(5250)} \approx 1.35$ m. Answer: the maximum compression is about 1.35 m, less than the 2.00 m found for the frictionless case, exactly as expected since friction also absorbs energy. Timing note: about three minutes. The trap is reaching for conservation of mechanical energy out of habit — the moment friction is mentioned, only the work-energy theorem, with friction's work as a separate term, gives the right equation. Keep every term's sign consistent with opposing the compression, and discard the negative root without a second thought.

Competitive level - glancing elastic collision of equal masses (JEE/NEET pattern)

Two billiard balls of equal mass $m_1=m_2$ undergo an elastic collision; the first ball (the cue ball) strikes the second (the target ball), initially at rest, and the target ball moves off at $\theta_2=37°$ from the cue ball's original direction. Find the angle $\theta_1$ at which the cue ball moves off, assuming friction and rotation are negligible. (Following NCERT Example 5.12.)

Step 1. Write momentum conservation as a vector equation: $\vec{v}_{1i} = \vec{v}_{1f}+\vec{v}_{2f}$ (masses cancel since they are equal). Square both sides by dotting each side with itself: $v_{1i}^2 = (\vec{v}_{1f}+\vec{v}_{2f})\cdot(\vec{v}_{1f}+\vec{v}_{2f}) = v_{1f}^2+v_{2f}^2+2v_{1f}v_{2f}\cos(\theta_1+\theta_2)$, using the fact that the angle between the two final velocity vectors is $\theta_1+\theta_2$ (they are measured on opposite sides of the original direction).

Step 2. Write kinetic energy conservation, valid since the collision is elastic and the masses are equal: $v_{1i}^2 = v_{1f}^2+v_{2f}^2$.

Step 3. Compare the two boxed expressions for $v_{1i}^2$ from Steps 1 and 2. Both equal $v_{1f}^2+v_{2f}^2$ plus an extra term in Step 1, so that extra term must vanish: $2v_{1f}v_{2f}\cos(\theta_1+\theta_2) = 0$. Since neither final speed is zero for a genuine glancing collision, $\cos(\theta_1+\theta_2)=0$, which forces $\theta_1+\theta_2=90°$.

Step 4. Substitute the given $\theta_2=37°$: $\theta_1 = 90°-37° = 53°$. Answer: $\theta_1=53°$, and more generally, two equal masses that undergo a glancing elastic collision (one initially at rest) always separate at right angles to each other — a result billiards and carrom players use without ever writing the algebra. Timing note: under two minutes once the equal-masses, elastic, one-at-rest pattern is recognised. This is a heavily re-used JEE and NEET result; memorise the conclusion, angle sum 90°, as a shortcut, but be ready to derive it since the given angle can change.

Zero force or zero displacement is not the only way to get zero work — the angle matters just as much, and students routinely forget that a large force acting over a large displacement can still do exactly zero work if the two are perpendicular, as with gravity on a body moving horizontally or the tension in a pendulum string, which never does work because it is always normal to the bob's motion. A second frequent slip is assuming that Newton's third law, which makes the forces between two bodies equal and opposite, also makes the work done by each on the other equal and opposite — it does not, as the skidding-cyclist example shows, where the road does negative work on the cycle but the cycle does zero work on the road. Sign errors plague spring problems: work done by the spring during a stretch is negative, while work done against it (stored as potential energy) is positive — mixing these up flips the final answer. Students also confuse average and instantaneous power, using $P=Fv$ with an average speed when the problem wants $P_{av}=W/t$, or vice versa. In natural-log calculations remember $\ln x = \log_e x = 2.303\log_{10}x$, not $\log_{10}x$ directly. Most importantly for collisions: momentum conservation applies to every collision without exception, but kinetic energy conservation applies only to elastic collisions — using the wrong pair of equations is the single most common error in this part of the chapter.

• Compute the dot product of two vectors both from magnitudes and angle, $\vec{A}\cdot\vec{B}=AB\cos\theta$, and from components, and use it to find the work done by a constant force.
• Identify, from physical reasoning alone, the three situations in which a force does zero work: zero displacement, zero force, or force perpendicular to displacement.
• State and apply the work-energy theorem, $K_f-K_i=W$, for both constant and variable forces, including cases where the exact force law is unknown.
• Compute the work done by a variable force as the area under a force-displacement graph, or by direct integration, $W=\int F(x)\,dx$.
• Define potential energy from a conservative force via $F=-dV/dx$, and derive $V(h)=mgh$ for gravity and $V(x)=\tfrac12kx^2$ for a spring.
• State and apply conservation of mechanical energy for conservative forces, and correctly modify it with $\Delta E=W_{nc}$ when friction or another non-conservative force is present.
• Calculate average and instantaneous power, convert between watts and horsepower, and distinguish power (a rate) from energy (a kilowatt-hour).
• Distinguish elastic, inelastic and completely inelastic collisions by what happens to kinetic energy, while recognising momentum is conserved in all three.
• Derive and apply the one-dimensional elastic collision formulas for $v_{1f}$ and $v_{2f}$, and recall the equal-mass and heavy-target special cases without re-deriving them each time.
• Set up and solve the momentum-and-energy equations for a two-dimensional (glancing) elastic collision, including the equal-mass right-angle result.

Ask the tutor: Stuck on when to use energy conservation versus the work-energy theorem with friction, or want a walkthrough of a two-dimensional collision problem — ask the tutor to work through it with you step by step.