Work, Energy and Power

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 5.

A weightlifter holds 150 kg for 30 seconds. How much work does he do?

None. In physics terms, not a single joule, however tired he feels. A satellite circling the Earth for years does no work either. Yet a falling stone, a braking car and a compressed spring all do work. This chapter gives "work" a precise meaning, and that meaning turns out to be the key to almost every mechanics problem: instead of tracking forces moment by moment, you track energy from start to finish.

Work: only the push along the motion counts

Push a sled with a rope held at an angle. The part of your pull that points forward speeds it up. The part that lifts the rope upward does nothing for the motion. Work captures exactly this: it is the component of force along the displacement, multiplied by the displacement.

$$W = F \cdot d = F d \cos \theta$$

Here θ is the angle between the force and the displacement. The symbol "·" is the scalar (dot) product of two vectors, $A \cdot B = AB\cos\theta$. It gives a number, not a direction, which is why work is a scalar. The dot product is commutative ($A \cdot B = B \cdot A$) and distributive over addition. For unit vectors, $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$. So in component form:

$$A \cdot B = A_x B_x + A_y B_y + A_z B_z, \qquad A \cdot A = A^2$$

Work can be positive, zero or negative depending on the angle, and the sign tells you who is giving energy to whom.

So work is zero in three situations: the displacement is zero (pushing a wall, holding a weight), the force is zero, or the force is perpendicular to the displacement. The SI unit is the joule (J), with dimensions $[M\,L^2\,T^{-2}]$.

Worked example: pulling a sled. A rope pulls a sled with a force of 50 N at 37° above the horizontal. The sled moves 10 m along the ground. How much work does the rope do? (Take $\cos 37^\circ = 0.8$.)

Common wrong step. Writing W = F d = 500 J. That would be right only if the rope were horizontal. The vertical part of the pull, $50 \sin 37^\circ = 30$ N, lifts the rope's end but contributes nothing to the sled's forward motion.

Pause and predict: a car brakes to a stop on a level road. The road pushes the car backwards with friction. Does the car do work on the road?

Check your answer: no. Newton's third law gives equal and opposite forces, but work needs displacement of the point of application. The road does negative work on the car, and the car does zero work on the road because the road does not move. The works done by an action-reaction pair need not be equal and opposite.

Kinetic energy and the work-energy theorem

Where does the half in $1/2\,mv^2$ come from? Start from the kinematic result $v^2 - u^2 = 2ad$. Multiply both sides by $m/2$:

$$\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2 = (ma)\,d = Fd$$

The right side is the work done by the net force. The left side is the change in the quantity $1/2\,mv^2$, so we give that quantity a name, kinetic energy:

$K = 1/2\,mv^2$, and Work-energy theorem: $K(\text{final}) - K(\text{initial}) = W(\text{net})$

Kinetic energy is a scalar and never negative. Because it depends on $v^2$, doubling the speed quadruples the energy. This explains road safety rules: a car at 80 km/h needs four times the braking distance of the same car at 40 km/h, because the brakes must remove four times the energy with the same force.

The theorem is also true for variable forces. Starting from $dK/dt = Fv = F\,(dx/dt)$, integrating over x gives $K(\text{final}) - K(\text{initial}) = \int F\,dx$. This is why the theorem is so powerful: it ignores the details of the path and the time, and looks only at the net work.

Beyond the textbook. For the exam: kinetic energy is $1/2\,mv^2$ and the work-energy theorem gives the speed change from the net work. In reality: the same speed-squared dependence is why vehicle crashes get so much worse with speed, and why a car's brakes get hot. The kinetic energy removed does not vanish. It appears as heat in the brake discs and tyres. Why it matters: "where did the energy go?" is the question that decides whether a CUET problem needs conservation of energy or the work-energy theorem with a friction term.

Variable force: work is the area under the graph

If the force changes with position, split the path into tiny steps Δx. On each step the force is nearly constant, so $\Delta W = F(x)\,\Delta x$. Adding up and letting $\Delta x \to 0$:

$W = \int F(x)\,dx$ from $x_1$ to $x_2$, which equals the area under the F–x graph

For a force-position graph made of triangles and rectangles you do not even need calculus. Count the areas, with areas below the axis counted as negative.

Worked example: a force that grows with distance. A particle is pushed by $F = (7 - 2x + 3x^2)$ N, where x is in metres. Find the work done from x = 0 to x = 5 m.

Common wrong step. Evaluating F at the end point (7 − 10 + 75 = 72 N) and multiplying by 5 m to get 360 J. The force was much smaller for most of the journey, so the end value overstates the work.

Potential energy and conservative forces

Lift a book and hold it above the table. You gave it energy that it can return as kinetic energy when you let go. This stored energy of position or configuration is potential energy. Near the Earth's surface, with the ground as zero:

$V(h) = mgh$, and the force is recovered from $F = -dV/dh = -mg$

The minus sign means the force points toward lower potential energy. Gravity pulls a ball down because that is the way V decreases.

Not every force has a potential energy. Only a conservative force does, and it can be recognised in three equivalent ways: the work it does depends only on the end points, the work over any closed loop is zero, or it can be written as $F = -dV/dx$. Gravity and the spring force are conservative. Friction is not.

Why friction cannot have a potential energy. Slide a block from A to B along a straight path and then along a long curved path. The curved path is longer, so friction removes more energy. If the work depended on the path, "stored energy at B" would have no single value. Gravity does not behave like that: climb a hill by a steep or a gentle path, the gravitational work depends only on the height gained.

Only differences of potential energy matter, so you may place the zero wherever it suits you (ground for gravity, the natural length for a spring), provided you stay consistent through the problem.

Conservation of mechanical energy

If only conservative forces do work, kinetic energy lost is potential energy gained, and the sum stays fixed:

$K + V = \text{constant}$, that is, $\Delta K + \Delta V = 0$

A ball dropped from height H has energy mgH at the top and $1/2\,mv^2$ at the bottom, so $mgH = 1/2\,mv^2$ and the landing speed is $v = \sqrt{2gH}$. At an intermediate height h the speed is $v = \sqrt{2g(H - h)}$. The result does not depend on the mass, and it is the same as the kinematic result, but the energy route also works on curved paths where kinematics fails.

Beyond the textbook. For the exam: mgh is the potential energy and a ball dropped from H lands at $\sqrt{2gH}$. In reality: mgh is an approximation that assumes g is constant. For large heights the exact expression is $-GMm/r$, which you meet in Gravitation. Air drag also removes some energy from every real fall, which is why a skydiver reaches a steady terminal speed instead of the $\sqrt{2gH}$ the formula predicts.

The vertical circle: a classic energy-plus-force problem

A bob of mass m on a light string of length L is whirled in a vertical circle. At the top the string tension T and the weight both point down, and together they provide the centripetal force: $T + mg = mv^2/L$. The bob can just complete the circle when T drops to zero at the top, which gives the minimum speed at the top, $v(\text{top}) = \sqrt{gL}$.

Now use energy. From the bottom to the top the bob rises 2L, so $1/2\,mv(\text{bottom})^2 = 1/2\,mv(\text{top})^2 + mg(2L)$. Substitute $v(\text{top})^2 = gL$:

$v(\text{bottom}) = \sqrt{5gL}$, the minimum speed at the lowest point

For L = 0.9 m and g = 10 m s⁻², $v(\text{top}) = 3$ m s⁻¹ and $v(\text{bottom}) = \sqrt{45} \approx 6.7$ m s⁻¹.

Common wrong step. Taking v(top) = 0 because "the bob stops at the highest point". That is true for a bob on a rigid rod, which can push as well as pull. A string can only pull, so the bob must still be moving fast enough at the top to keep the string taut.

The spring: a force that grows as you stretch it

A spring obeys Hooke's law, $F = -kx$, where k is the spring constant (N m⁻¹) and the minus sign says the force opposes the displacement. The force changes with x, so the work done in stretching it from 0 to x is the area of the triangle under the F–x graph:

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

Stretching and compressing by the same amount store the same energy because x is squared. Doubling the compression quadruples the energy. If a block of mass m is released from a compression x on a smooth surface, the energy $1/2\,kx^2$ turns into $1/2\,mv^2$. The speed is largest at the natural length, where the potential energy is zero, and equals $v(\text{max}) = x\sqrt{k/m}$.

Worked example: a released spring. A 2 kg block compresses a spring (k = 200 N m⁻¹) by 0.30 m on a smooth table and is then released. Find the energy stored and the speed with which the block leaves the spring.

Common wrong step. Computing the energy as F × x = (200 × 0.30) × 0.30 = 18 J. The spring force is not 60 N during the whole compression. It rises from 0 to 60 N, so the average is 30 N and the work is 9 J. That is the ½.

Pause and predict: if the same block is compressed twice as far (0.60 m), by what factor does the exit speed change?

Check your answer: it doubles to 6 m s⁻¹. The energy rises by 4 ($x^2$), and v goes as the square root of energy, so v is proportional to x.

Energy with friction

When a non-conservative force acts, mechanical energy is no longer conserved. The change in mechanical energy equals the work of the non-conservative force:

$E(\text{final}) - E(\text{initial}) = W(\text{non-conservative})$

Worked example: a rough slope. A 2 kg block slides from rest down a rough incline and drops through a height of 3 m. It reaches the bottom at 6 m s⁻¹. Take g = 10 m s⁻². How much work does friction do?

So 24 J of mechanical energy was converted to heat. Common wrong step. Using $v = \sqrt{2gh} = 7.7$ m s⁻¹, which ignores friction entirely. The actual speed is lower exactly because friction took 24 J.

Power: how fast energy is delivered

Two cranes can lift the same load, but the faster one has more power. Power is the rate of doing work.

$P(\text{average}) = W/t$, and $P = dW/dt = F \cdot v$

Power is a scalar with SI unit watt (1 W = 1 J s⁻¹) and dimensions $[M\,L^2\,T^{-3}]$. Two other units appear on CUET: 1 horsepower = 746 W, and the kilowatt-hour, 1 kWh = $3.6 \times 10^6$ J. The kilowatt-hour is a unit of energy, because it is power multiplied by time. It is what your electricity bill counts as "one unit".

The form $P = Fv$ is the useful one for vehicles: at constant speed the engine force equals the resistive force, so $P = (\text{resistance}) \times v$.

Worked example: engine power. A car moves at a steady 72 km h⁻¹ on a level road against a total resistance of 1500 N. What power does the engine deliver?

Common wrong step. Multiplying by 72 instead of 20, which gives 108 kW. Whenever a formula uses SI units, kilometres per hour must be converted first.

Beyond the textbook. For the exam: if the resistance stays constant, doubling the speed doubles the power needed ($P = Fv$). In reality: air drag on a vehicle grows roughly as $v^2$, so the power to overcome it grows as $v^3$. Doubling the speed then needs about eight times the power. That is why highway fuel economy falls so sharply at high speed.

Collisions: momentum is the safe bet

In a collision the forces between the bodies are huge and last a very short time, so we don't track them. We use conservation laws instead. Total linear momentum is always conserved in a collision (the internal forces are equal and opposite, by Newton's third law). Kinetic energy is another matter.

The coefficient of restitution $e = (\text{relative speed of separation})/(\text{relative speed of approach})$. A ball dropped from height h that rebounds to h′ has $e = \sqrt{h'/h}$.

Perfectly inelastic, one body at rest. Mass $m_1$ moving at u hits $m_2$ at rest and they stick. Momentum gives $m_1u = (m_1 + m_2)v$:

$$v = \dfrac{m_1 u}{m_1 + m_2}, \qquad \text{kinetic energy lost} = \dfrac{1}{2} \cdot \dfrac{m_1 m_2 u^2}{m_1 + m_2}$$

The fraction of the initial kinetic energy that is lost is $m_2/(m_1 + m_2)$. It is never the whole kinetic energy, because the combined mass keeps moving.

Elastic, head-on, $m_2$ at rest. Conserving both momentum and kinetic energy (or, equivalently, using e = 1) gives:

$$v_1 = \dfrac{(m_1 - m_2)u}{m_1 + m_2}, \qquad v_2 = \dfrac{2m_1 u}{m_1 + m_2}$$

Three special cases are worth remembering. If $m_1 = m_2$, the first body stops and the second moves off with speed u (the velocities are exchanged, as in Newton's cradle). If $m_1 \ll m_2$ (a ball hitting a wall), the light body rebounds with nearly its own speed, $v_1 \approx -u$, and the heavy body hardly moves. If $m_1 \gg m_2$ (a truck hitting a ball), the truck is barely slowed and the ball flies off at about 2u. A negative $v_1$ simply means the first body rebounds.

For example, a 2 kg body at 6 m s⁻¹ strikes a 4 kg body at rest elastically. Then $v_1 = (2 - 4)(6)/6 = -2$ m s⁻¹ and $v_2 = 2(2)(6)/6 = 4$ m s⁻¹. Momentum before is 12 kg m s⁻¹ and after is −4 + 16 = 12. Energy before is 36 J and after is 4 + 32 = 36 J.

Worked example: bullet and hanging block. A 20 g bullet moving at 300 m s⁻¹ embeds in a 1.98 kg block hanging from a long string. How high does the block swing, and how much energy is lost in the impact? Take g = 10 m s⁻².

Common wrong step. Equating the bullet's initial energy to the block's potential energy, which gives a height of 45 m. Energy conservation holds only after the collision. During the impact almost all the energy goes into heating and deforming the block and the bullet. The shortcut: the fraction lost is $m_2/(m_1 + m_2) = 1.98/2.00 = 0.99$.

Beyond the textbook. For the exam: in an elastic collision with $m_2$ at rest, the fraction of kinetic energy passed to $m_2$ is $4m_1m_2/(m_1 + m_2)^2$, which is largest when the masses are equal. In reality: this is why nuclear reactors use light materials such as water or graphite to slow neutrons. A neutron colliding with a particle of nearly its own mass gives up most of its energy in a single collision, while one bouncing off a heavy nucleus keeps almost all of it.

Collisions in two dimensions

If the bodies do not meet head-on, the motion spreads into a plane. Momentum is conserved separately along x and y. With $m_1$ moving along x and $m_2$ at rest, and after the collision the bodies going off at angles $\theta_1$ and $\theta_2$ on opposite sides of the x-axis:

$$m_1 u = m_1 v_1 \cos\theta_1 + m_2 v_2 \cos\theta_2, \quad \text{and} \quad 0 = m_1 v_1 \sin\theta_1 - m_2 v_2 \sin\theta_2$$

For an elastic collision the energy equation is a third equation. Still, with four unknowns ($v_1$, $v_2$, $\theta_1$, $\theta_2$), one of them has to be given.

Why equal masses fly apart at 90°. In an elastic glancing collision between equal masses with one at rest, momentum gives the vector equation $u = v_1 + v_2$, and energy gives $u^2 = v_1^2 + v_2^2$. Squaring the first, $u^2 = v_1^2 + v_2^2 + 2\,v_1 \cdot v_2$. Comparing the two, $v_1 \cdot v_2 = 0$, so the final velocities are perpendicular. Billiard players use this every day.

Energy has many forms

The broadest statement is that the total energy of an isolated system is constant: mechanical, heat, electrical, chemical and nuclear energy convert into one another. Mechanical energy is conserved only when conservative forces alone do the work. In 1905 Einstein added that mass itself is a form of energy, $E = mc^2$. One gram of mass is equivalent to $9 \times 10^{13}$ J. Chemical reactions change mass by an unmeasurably tiny amount, so in mechanics we ignore it. In nuclear fission and fusion the change is large enough to matter, as you will see in the chapter on Nuclei.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Laws of Motion supplies the forces and the momentum conservation used in collisions.
• Next: System of Particles and Rotational Motion rebuilds work, power and kinetic energy for rotation ($\tau\,d\theta$, $\tau\omega$ and $1/2\,I\omega^2$).
• Same ideas later: Gravitation replaces mgh with $-GMm/r$, and Oscillations uses the spring energy $1/2\,kx^2$ to describe simple harmonic motion.
• Class XII link: Current Electricity reuses power as $P = VI$, and Nuclei uses $E = mc^2$ for binding energy.