Work, Energy and Power: Full Notes

5.1 Work Done by a Constant and a Variable Force

In everyday speech, “work” means almost any kind of effort, holding up a heavy bag tires your arm, so surely that counts? In physics it does not. Work is done only when a force actually moves something along its own direction. Holding a bag still does zero work on it, however tired your arm gets, because the bag does not move. Walking forward with that same bag at constant height also does zero work against gravity, because your vertical lifting force and the bag’s horizontal displacement are perpendicular to each other.

For a constant force $\mathbf{F}$ acting while a body undergoes a displacement $\mathbf{d}$ , work is defined as the dot product of the two:

W = \mathbf{F}\cdot\mathbf{d} = Fd\cos\theta

where $\theta$ is the angle between the force and the displacement. Only $F\cos\theta$ , the component of the force along the displacement, ever contributes. This is why the single formula quietly covers three cases at once: work is positive when the force has a component in the direction of motion ( $\theta<90^\circ$ , gravity doing work on a falling stone), negative when it opposes the motion ( $\theta=180^\circ$ , friction on a sliding block, or gravity on a rising ball), and exactly zero when the force is perpendicular to the motion ( $\theta=90^\circ$ , the tension in a conical pendulum’s string, or the normal force on anything moving along a level surface).

A push of 150 N dragging a 20 kg suitcase at 30° above the floor across 10 m does work $W = 150\times10\times\cos30^\circ \approx 1299\text{ J}$ on the suitcase, regardless of how heavy it is, only the applied force, the distance, and the angle matter.

5.1.1 Work by a variable force

Many real forces are not constant. A spring pulls back harder the further you stretch it; a rocket’s weight changes as it burns fuel. For a force that varies with position, the trick is to chop the path into tiny steps $\Delta x$ short enough that $F$ is practically constant over each one, add up the little bits of work $F(x)\,\Delta x$ , and let the steps shrink to zero. That sum becomes an integral, and geometrically it is simply the area under the F–x graph:

W = \int_{x_i}^{x_f} F(x)\,dx

A spring stretched by $x$ from its natural length pulls back with $F=-kx$ (Hooke’s law, with $k$ the spring constant), so the external agent stretching it must supply $+kx$ at every instant. Plotted against $x$ , that external force is a straight line through the origin, and the work done in stretching it from $0$ to $x$ is just the area of the triangle under that line, base $x$ , height $kx$ :

Work to stretch a spring

W = \dfrac{1}{2}kx^2

Worked example

Stretching a spring

A spring of force constant $k = 200\text{ N/m}$ is stretched by 5 cm from its natural length. Find the work done.

Converting 5 cm to 0.05 m and substituting directly:

W = \dfrac{1}{2}kx^2 = \dfrac{1}{2}\times200\times(0.05)^2 = \textcolor{#e08a1e}{0.25\text{ J}}

As a check, the triangle’s peak force is $F_{max}=kx=200\times0.05=10\text{ N}$ , so its area is $\tfrac12\times0.05\times10 = 0.25\text{ J}$ , the same answer either way.

5.2 Kinetic Energy and the Work-Energy Theorem

Kinetic energy is the energy a body has by virtue of its motion, and work is precisely the mechanism that creates or removes it. Consider a constant force $F=ma$ acting alone on a mass $m$ , pushing its speed from $u$ to $v$ over a displacement $s$ . Kinematics gives $v^2=u^2+2as$ , so $as=\tfrac12(v^2-u^2)$ , and the work done is:

W = Fs = mas = \tfrac12 mv^2 - \tfrac12 mu^2

The quantity $\tfrac12 mv^2$ that appears on both ends is what we call kinetic energy, $K=\tfrac12 mv^2$ . The same result drops out of calculus without ever assuming $F$ is constant: since $dW = F\,dx = m\dfrac{dv}{dt}dx = mv\,dv$ , integrating from 0 to $v$ gives $W=\int_0^v mv\,dv = \tfrac12 mv^2$ directly, confirming this holds for a variable force too. Either way, we arrive at the work-energy theorem:

Work-Energy Theorem

\textcolor{#e08a1e}{W_{net}} = K_f - K_i = \Delta K

the total (net) work done by all forces on a body equals its change in kinetic energy. This is really just Newton’s second law rewritten in terms of energy rather than force, which makes it enormously convenient whenever you only care about speeds at two points and not about the messy details of the motion in between.

Worked example

Braking distance from the work-energy theorem

A scooter with its rider has a total mass of 90 kg and is moving at 36 km/h when the brakes are applied, bringing it to rest over 25 m. Find the average braking force.

First convert the speed: 36 km/h = 10 m/s. The initial kinetic energy:

K_i = \tfrac12\times90\times10^2 = 4500\text{ J}

Since the scooter ends at rest, $K_f=0$ , so the net work done by the brakes (the only horizontal force) is $-4500\text{ J}$ . Writing that as $-Fd$ :

-F\times25 = -4500 \;\Rightarrow\; F = \textcolor{#e08a1e}{180\text{ N}}

Notice we never needed the braking time or the exact way the speed fell, only the two kinetic energies and the distance.

5.3 Potential Energy and Conservative Forces

Potential energy is energy stored in the configuration of a system, the stretch of a spring, or the height of a raised stone, released as kinetic energy when that configuration is allowed to relax. It can only be defined for a special class of forces called conservative forces: a force is conservative if the work it does on a body moving between two points does not depend on the path taken, only on the two endpoints. Equivalently, the work done by a conservative force around any closed loop, out and back to the same point, is exactly zero.

Gravity is the cleanest example: lifting a block of mass $m$ to a height $h$ takes work $mgh$ against gravity, whether you go straight up, diagonally, or by the most roundabout staircase imaginable, the horizontal detours contribute nothing because gravity is vertical. Defining potential energy as the negative of the work done by the force, and setting $U=0$ at the ground, gives the familiar:

Gravitational PE

U(h) = mgh

Friction is the opposite story. Slide a crate 2 m in a straight line across a rough floor and friction does a certain amount of negative work; drag it the same net distance by an unnecessary 10 m detour and friction does five times as much negative work, converted entirely into heat at the sliding surfaces. Because that work depends on the length of the path and not just the endpoints, friction is non-conservative, and more importantly it is dissipative: once that energy becomes heat, it cannot spontaneously reassemble itself back into the crate’s motion. This single distinction, path-independent and recoverable versus path-dependent and lost as heat, is exactly what separates the ramp problems in the next section into “mechanical energy conserved” and “mechanical energy visibly leaking away”.

5.4 The Potential Energy of a Spring and Conservation of Mechanical Energy

A spring stores the work done in deforming it as elastic potential energy, following directly from §5.1’s triangle-area result with $x$ now read as the spring’s displacement from its natural length:

Spring PE

U(x) = \dfrac{1}{2}kx^2

Now combine §5.2’s theorem with §5.3’s definition of a conservative force. The work-energy theorem says $W_{net}=\Delta K$ , and for a conservative force the work it does is minus the change in its potential energy, $W_{cons} = -\Delta U$ . If conservative forces are the only forces doing work (no friction, no applied force, nothing external), then $W_{net}=W_{cons}$ , and:

\Delta K = -\Delta U \;\;\Longrightarrow\;\; K + U = \text{constant}

This is the law of conservation of mechanical energy: a dropped stone trades PE for KE at exactly a one-for-one rate as it falls, and a block on a frictionless spring trades KE for spring PE and back, oscillating forever with $K+U$ never changing.

5.4.1 When friction is added: mechanical energy versus total energy

Put the same block on a rough incline of length $L$ , angle $\theta$ , with friction coefficient $\mu$ between block and surface, exactly the setup in the interactive ramp simulation below. Friction now does real negative work, so mechanical energy is no longer constant, it visibly shrinks. But nothing is actually lost: every joule of mechanical energy that disappears reappears as heat at the sliding surface, so the total energy, mechanical plus heat, is still conserved.

First, whether the block moves at all depends on §4.7’s static friction ceiling: it starts sliding only if $mg\sin\theta$ (the pull along the slope) exceeds the maximum friction $\mu mg\cos\theta$ , i.e. only if $\tan\theta > \mu$ . When it does slide, taking the bottom of the incline as the zero of height, the total gravitational PE available over the whole slide is:

U_0 = mgL\sin\theta

After the block has slid a distance $s$ down the slope, its height above the bottom is $(L-s)\sin\theta$ , so its remaining PE and the heat generated so far (friction $\mu mg\cos\theta$ acting over distance $s$ ) are:

U(s) = mg(L-s)\sin\theta \qquad\qquad Q(s) = \mu mg\cos\theta \cdot s

and since nothing but gravity and friction acts, the work-energy theorem guarantees that whatever PE is missing has gone into KE and heat together:

K(s) = U_0 - U(s) - Q(s)

Drag the simulation’s sliders and watch the blue (PE), light-blue (KE), and orange (heat) bar shift in real time, the blue shrinks, but blue + light-blue + orange always adds back up to the same total, $U_0$ , confirming that it is total energy, not mechanical energy, that is truly conserved.

Worked example

Speed at the bottom of a rough incline

A 2 kg block starts from rest at the top of a 4 m incline tilted at 30°, with $\mu=0.2$ between block and incline. Use energy methods to find its speed at the bottom. ( $g=9.8\text{ m/s}^2$ )

Since $\tan30^\circ \approx 0.577 > 0.2$ , the block does slide. The total PE available, taking the full length $s=L=4\text{ m}$ at the bottom (height above bottom is then zero):

U_0 = mgL\sin\theta = 2\times9.8\times4\times\sin30^\circ = 39.2\text{ J}

The heat generated over the full slide:

Q = \mu mg\cos\theta \cdot L = 0.2\times2\times9.8\times\cos30^\circ\times4 \approx 13.58\text{ J}

So the kinetic energy at the bottom (where the remaining PE is zero):

K = U_0 - 0 - Q = 39.2 - 13.58 = 25.62\text{ J}

\tfrac12 mv^2 = 25.62 \;\Rightarrow\; v = \sqrt{\dfrac{2\times25.62}{2}} \approx \textcolor{#e08a1e}{5.06\text{ m/s}}

which matches the purely kinematic route, $a = g(\sin\theta-\mu\cos\theta)\approx3.20\text{ m/s}^2$ , giving $v=\sqrt{2aL}\approx5.06\text{ m/s}$ as well, exactly as it must.

5.5 Power

Energy and work tell you how much gets done; power tells you how fast. Two motors can lift the same weight through the same height, doing identical work, yet one might take a second and the other a minute, a difference that matters enormously in engineering even though it is invisible to the work-energy theorem alone. Average power is simply work over time:

P_{avg} = \dfrac{W}{t}

and instantaneous power is its limit as $t\to0$ . Since $dW = \mathbf{F}\cdot d\mathbf{x}$ , dividing by $dt$ gives a neat second formula entirely in terms of the instantaneous velocity:

Instantaneous Power

P = \dfrac{dW}{dt} = \mathbf{F}\cdot\mathbf{v}

The SI unit is the watt (1 W = 1 J/s), named for James Watt, who needed a unit to advertise how many draft horses his steam engines could replace, which is exactly where the older unit horsepower (1 hp ≈ 746 W) comes from.

Worked example

Average power climbing a rope

A 50 kg boy climbs a 10 m vertical rope at a steady pace, reaching the top in 20 s. Find his average power output. ( $g=9.8\text{ m/s}^2$ )

The work done is entirely against gravity, raising his weight through 10 m:

W = mgh = 50\times9.8\times10 = 4900\text{ J}

P_{avg} = \dfrac{W}{t} = \dfrac{4900}{20} = \textcolor{#e08a1e}{245\text{ W}}

about a third of a horsepower, sustained for the full climb, a reasonable output for a short, hard effort.

5.6 Collisions

A collision is any brief, intense interaction between two bodies. Whatever mutual forces act during the collision, they are internal to the two-body system and, by Newton’s third law (§4.4), always cancel in pairs, so total momentum is conserved in every collision, elastic or not, exactly as in §4.5. What is not automatically conserved is kinetic energy. A collision is called elastic if kinetic energy happens to come out unchanged (idealized billiard balls, gas molecules), and inelastic if some of it is converted to heat, sound, or permanent deformation, with a perfectly inelastic collision being the extreme case where the bodies stick together and move off with one common velocity, losing the maximum kinetic energy momentum conservation allows.

5.6.1 Elastic collisions in one dimension

For two bodies of mass $m_1,m_2$ moving along the same line with initial velocities $u_1,u_2$ and final velocities $v_1,v_2$ , both momentum and kinetic energy are conserved:

m_1u_1+m_2u_2 = m_1v_1+m_2v_2

\tfrac12 m_1u_1^2+\tfrac12 m_2u_2^2 = \tfrac12 m_1v_1^2+\tfrac12 m_2v_2^2

Rearranging the momentum equation as $m_1(u_1-v_1)=m_2(v_2-u_2)$ and the energy equation (after cancelling the halves and factoring each side as a difference of squares) as $m_1(u_1-v_1)(u_1+v_1)=m_2(v_2-u_2)(v_2+u_2)$ , dividing one by the other collapses the quadratic mess into a clean linear condition: $u_1+v_1=u_2+v_2$ , i.e. the relative velocity of separation equals the relative velocity of approach. Solving that together with momentum conservation gives the final velocities for general masses:

v_1 = \dfrac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2} \qquad v_2 = \dfrac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2}

A nice special case is equal masses, $m_1=m_2$ : the formulas collapse to $v_1=u_2$ and $v_2=u_1$ , the two bodies simply swap velocities. A cue ball striking an identical stationary ball dead-on stops completely while the struck ball shoots off with the cue ball’s entire original speed, the classic pool-table shot.

Worked example

Elastic collision between unequal masses

A 4 kg ball moving at 5 m/s strikes a stationary 1 kg ball head-on, elastically. Find both velocities after the collision.

Here $m_1=4,\ u_1=5,\ m_2=1,\ u_2=0$ . Substituting into the two formulas:

v_1 = \dfrac{(4-1)\times5 + 0}{5} = \dfrac{15}{5} = \textcolor{#e08a1e}{3\text{ m/s}}

v_2 = \dfrac{0 + 2\times4\times5}{5} = \dfrac{40}{5} = \textcolor{#e08a1e}{8\text{ m/s}}

As a check, momentum before and after both equal 20 kg m/s:

4\times5 = 20 \qquad 4\times3+1\times8 = 12+8=20 \;\checkmark

and kinetic energy before and after both equal 50 J:

\tfrac12(4)(5)^2=50 \qquad \tfrac12(4)(3)^2+\tfrac12(1)(8)^2 = 18+32=50\ \checkmark

— From the NCERT exercises

A couple of the chapter’s own exercise questions, with original worked solutions.

NCERT Exercise 5.12

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 minutes. If the tank is 40 m above the ground and the efficiency of the pump is 30%, how much electric power is consumed by the pump? (Take g = 9.8 m/s², density of water = 1000 kg/m³.)

Solution

The mass of water moved is $1000\times30 = 30{,}000\text{ kg}$ , lifted through 40 m, so the useful work output and the time taken are:

W = mgh = 30000\times9.8\times40 = 1.176\times10^{7}\text{ J}

t = 15\text{ min} = 900\text{ s}

which gives the pump’s useful (output) power:

P_{output} = \dfrac{W}{t} = \dfrac{1.176\times10^{7}}{900} \approx 1.307\times10^{4}\text{ W}

Only 30% of the electrical power actually goes into lifting water, so the electric power consumed is larger than this by a factor of $1/0.30$ :

P_{input} = \dfrac{P_{output}}{0.30} = \dfrac{1.307\times10^{4}}{0.30} \approx \textcolor{#e08a1e}{4.36\times10^{4}\text{ W}}

NCERT Exercise 5.33

A bullet of mass 0.012 kg and horizontal speed 70 m/s strikes a block of wood of mass 0.4 kg, initially at rest on a frictionless table, and gets embedded in it instantly. Find the common speed acquired, and calculate the amount of heat produced in the collision.

Solution

This is a perfectly inelastic collision: momentum is conserved, but kinetic energy is not. Conserving momentum, with the bullet-block system moving at common speed $v$ afterward:

0.012\times70 = (0.012+0.4)\,v \;\Rightarrow\; v = \dfrac{0.84}{0.412} \approx \textcolor{#e08a1e}{2.04\text{ m/s}}

The kinetic energy before the collision belongs entirely to the bullet:

K_i = \tfrac12\times0.012\times70^2 = 29.4\text{ J}

and after the collision, the combined mass moves at the common speed found above:

K_f = \tfrac12\times0.412\times(2.04)^2 \approx 0.857\text{ J}

so the heat produced, the kinetic energy that simply disappears, is:

Q = K_i-K_f \approx 29.4-0.857 \approx \textcolor{#e08a1e}{28.5\text{ J}}

Almost all of the bullet’s original kinetic energy (about 97% of it) is lost as heat and deformation in the wood, only a sliver survives as the kinetic energy of the combined, much heavier, slower-moving block.