Work, Energy and Power

In this chapter: scalar product, work by constant and variable forces, kinetic energy, the work-energy theorem, conservative forces and potential energy, conservation of mechanical energy, spring potential energy, the vertical circle, power, and elastic and inelastic collisions.

The scalar product

The scalar (dot) product of two vectors is A · B = AB cos θ, where θ is the angle between them. The result is a scalar. It is commutative and distributive, and î · î = ĵ · ĵ = k̂ · k̂ = 1 while î · ĵ = ĵ · k̂ = k̂ · î = 0. In component form, A · B = A_xB_x + A_yB_y + A_zB_z. Two non-zero vectors are perpendicular if their dot product is zero.

Work

The work done by a constant force F during a displacement d is

• Positive work for 0 ≤ θ < 90°: gravity on a falling ball.
• Negative work for 90° < θ ≤ 180°: kinetic friction on a sliding block, gravity on a ball going up.
• Zero work when the displacement is zero (pushing a wall) or θ = 90° (the normal force on a block sliding on a floor, the centripetal force in uniform circular motion).

Work done by a variable force

If the force varies with position, break the displacement into small bits and add: W = ∫_xi^x_f F(x) dx. On a force-displacement graph this is the area under the curve, with area below the x-axis counting as negative work.

Kinetic energy and the work-energy theorem

Kinetic energy is K = ½mv² = p²/2m, a scalar that is never negative.

Worked example: A raindrop of mass 1.00 g falls from a height of 1.00 km and hits the ground with a speed of 50.0 m s⁻¹. Find the work done by gravity and by the resistive force of air. (g = 10 m s⁻²)
Solution: Work by gravity = mgh = 10⁻³ × 10 × 10³ = 10.0 J. Final KE = ½ × 10⁻³ × (50)² = 1.25 J, starting from rest. By the work-energy theorem, W_g + W_r = 1.25 J, so W_r = 1.25 − 10.0 = −8.75 J.

Potential energy and conservative forces

A force is conservative if the work it does depends only on the initial and final positions, not on the path. Equivalently, the work done by it around any closed path is zero. Gravity and the spring force are conservative; friction and air drag are not.

For a conservative force we can define a potential energy V such that the work done by the force equals the decrease in V:

Conservation of mechanical energy

If only conservative forces do work, the total mechanical energy K + V stays constant. When non-conservative forces like friction act, the mechanical energy decreases by exactly the work they do against the motion, and that energy appears as heat. A body released from rest at height H has speed √(2gH) on reaching the ground, whatever path it takes along a smooth surface.

Potential energy of a spring

An ideal spring obeys Hooke's law, F_s = −kx, where k is the spring constant (N m⁻¹) and x is the extension or compression. The work done in stretching it from 0 to x is stored as

Worked example: A car of mass 1000 kg moving at 18 km h⁻¹ on a smooth road hits a horizontal spring of spring constant 6.25 × 10³ N m⁻¹. Find the maximum compression.
Solution: v = 18 × 5/18 = 5 m s⁻¹, so K = ½ × 1000 × 25 = 12 500 J. At maximum compression all of it is in the spring: ½kx_m² = 12 500, so x_m² = 25 000/6250 = 4 and x_m = 2 m.

Motion in a vertical circle

A bob on a string of length L is whirled in a vertical circle. Tension does no work (it is always perpendicular to the velocity), so mechanical energy is conserved. At the top, the string is just slack when the tension is zero and gravity alone provides the centripetal force: mg = mv²/L. Using energy conservation between the top and bottom (height difference 2L):

Power

Power is the rate of doing work.

The kilowatt hour on your electricity bill is a unit of energy, not power: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

Collisions

In any collision between two bodies with no external force, total linear momentum is conserved. Total energy is also conserved, but kinetic energy may not be, because some of it can become heat, sound or deformation.

• Elastic collision: both momentum and kinetic energy are conserved.
• Inelastic collision: momentum is conserved but some kinetic energy is lost.
• Perfectly inelastic collision: the bodies stick together after the collision. The loss of kinetic energy is the maximum possible.

Elastic collision in one dimension

Mass m₁ moving at u₁ hits mass m₂ at rest, head-on:

• Equal masses exchange velocities: the first stops and the second moves off with u₁. This is used to slow neutrons in a reactor with light nuclei.
• A very heavy body hitting a light one at rest hardly slows down, and the light one moves off at about 2u₁.
• A light body hitting a very heavy one bounces back with nearly the same speed.

In a perfectly inelastic collision with m₂ at rest, the common velocity is v = m₁u₁/(m₁ + m₂). In an elastic oblique collision of two equal masses with one at rest, the two move off at right angles to each other.

Worked example: A 2 kg ball moving at 3 m s⁻¹ collides with a 1 kg ball at rest and they stick together. Find the common velocity and the loss of kinetic energy.
Solution: 2 × 3 = (2 + 1)v, so v = 2 m s⁻¹. KE before = ½ × 2 × 9 = 9 J; KE after = ½ × 3 × 4 = 6 J. Loss = 3 J, one third of the initial KE.

Common mistakes: (1) Using only the applied force in the work-energy theorem; it needs the work of all forces, including friction and gravity. (2) Saying the work done by static friction is always zero; on a box accelerating on a truck floor it does positive work. (3) Assuming kinetic energy is conserved in every collision; only momentum is always conserved. (4) Writing V = mgh for large heights where g changes; use the gravitation formula instead. (5) Forgetting that stretching a spring from x to 2x needs three times the work of stretching it from 0 to x.

JEE and NEET focus

• Work from a force-displacement graph and work by a variable force using integration.
• Work-energy theorem with friction, and the relation K = p²/2m (percentage change problems).
• Spring problems: maximum compression, speed at a given extension.
• Vertical circle: critical speeds and tension at the top and bottom.
• One-dimensional elastic and perfectly inelastic collisions, fraction of KE lost.

Practice questions