Work, Energy & Power

Work is done when a force displaces a body.
For a constant force, W = F d cos($\theta$), where F is the force magnitude (N), d is the displacement (m), and $\theta$ is the angle between the force and displacement vectors. Dimensional formula: [$M^{1} \; L^{2} \; T^{-2}$], SI unit: joule (J). When $\theta$ < 90 degrees, cos($\theta$) > 0 and work is positive (force aids motion). When $\theta$ > 90 degrees, cos($\theta$) < 0 and work is negative (force opposes motion).
When $\theta$ = 90 degrees, cos($\theta$) = 0 and work is zero (force is perpendicular to displacement — as with centripetal force and normal force on a level surface).
For a variable force, W = integral of F dx, which equals the area under the F-x graph.

Kinetic energy KE = $\frac{1}{2}$ $mv^{2}$ [$M^{1} \; L^{2} \; T^{-2}$] (J), where m is mass (kg) and v is speed (m/s).
A useful alternative form is KE = $p^{2}$ / (2m), relating kinetic energy to momentum p.
The Work-Energy Theorem states that the net work done on a body equals its change in kinetic energy: $W_{net}$ = $\Delta$-KE = $\frac{1}{2}$ $mv^{2}$ - $\frac{1}{2}$ $\mu$². This includes work by ALL forces — gravity, friction, applied forces, etc.

Potential energy represents stored energy due to configuration.
Gravitational PE = mgh [$M^{1} \; L^{2} \; T^{-2}$] (J), where h is height above a chosen reference level.
Spring PE = $\frac{1}{2}$ $kx^{2}$, where k is the spring constant [$M^{1} \; L^{0} \; T^{-2}$] (N/m) and x is the extension or compression (m). Conservative forces (gravity, spring force) do work independent of path; non-conservative forces (friction, air resistance) depend on path.
When only conservative forces act, total mechanical energy is conserved: KE + PE = constant.

Power is the rate of doing work: P = W/t = F v cos($\theta$) [$M^{1} \; L^{2} \; T^{-3}$] (W, watt). Conversion: 1 horsepower (hp) = 746 W.

In vertical circular motion with a string: at the top, both tension T and weight mg point toward the centre, giving T + mg = $mv^{2}$/R, so minimum speed at the top is $v_{top}$ = $\sqrt{gR}$ (when T = 0).
At the bottom, T - mg = $mv^{2}$/R, so $T_{bottom}$ = $mv^{2}$/R + mg.
Using energy conservation from bottom to top: $v_{bottom}$ = $\sqrt{5gR}$.
For a rigid rod, the minimum speed at the top is zero (the rod can push), so $v_{bottom}$ = $\sqrt{4gR}$.

Collisions conserve momentum always.
In elastic collisions, kinetic energy is also conserved: $v_{1}$ = (($m_{1}$ - $m_{2}$)$u_{1}$ + $2m_{2} \; u_{2}$)/($m_{1}$ + $m_{2}$) and $v_{2}$ = (($m_{2}$ - $m_{1}$)$u_{2}$ + $2m_{1} \; u_{1}$)/($m_{1}$ + $m_{2}$). Special cases: equal masses exchange velocities; heavy body hitting light body at rest — light body moves at nearly 2u.
In perfectly inelastic collisions, bodies stick together: ($m_{1}$ + $m_{2}$)v = $m_{1} \; u_{1}$ + $m_{2} \; u_{2}$, with maximum KE loss.
The coefficient of restitution e = ($v_{2}$ - $v_{1}$)/($u_{1}$ - $u_{2}$): e = 1 (elastic), 0 < e < 1 (partially inelastic), e = 0 (perfectly inelastic).

Solved Numerical 1: A 2 kg block is pushed 5 m along a rough horizontal surface ($\mu_{k}$ = 0.3) by F = 20 N at 30 degrees to horizontal (g = 10 m/$s^{2}$).
Normal force: N = mg - F sin 30 = 2(10) - 20(0.5) = 20 - 10 = 10 N.
Friction: $f_{k}$ = $\mu_{k}$ N = 0.3 x 10 = 3 N.
Work by applied force: $W_{F}$ = F cos 30 x d = 20(0.866)(5) = 86.6 J.
Work by friction: $W_{f}$ = -$f_{k}$ d = -3(5) = -15 J.
Work by gravity: $W_{g}$ = 0 J (perpendicular).
Work by normal: $W_{N}$ = 0 J (perpendicular).
Net work: $W_{net}$ = 86.6 - 15 = 71.6 J.

Solved Numerical 2: Ball of mass m = 0.5 kg on string of length R = 2 m.
For complete vertical circular motion: Minimum speed at top: $v_{top}$ = $\sqrt{gR}$ = $\sqrt{10 x 2}$ = $\sqrt{20}$ = 4.47 m/s.
Minimum speed at bottom: $v_{bottom}$ = $\sqrt{5gR}$ = $\sqrt{5 x 10 x 2}$ = $\sqrt{100}$ = 10.0 m/s.

Solved Numerical 3: A 4 kg ball at $u_{1}$ = 6 m/s collides head-on elastically with a 2 kg ball at rest ($u_{2}$ = 0).
$v_{1}$ = ($m_{1}$ - $m_{2}$)$u_{1}$/($m_{1}$ + $m_{2}$) = (4 - 2)(6)/(4 + 2) = $\frac{12}{6}$ = 2 m/s.
$v_{2}$ = $2m_{1} \; u_{1}$/($m_{1}$ + $m_{2}$) = 2(4)(6)/(6) = 8 m/s.
Check momentum: 4(6) = 24 = 4(2) + 2(8) = 8 + 16 = 24 kg m/s.
Check KE: $\frac{1}{2}$(4)(36) = 72 J = $\frac{1}{2}$(4)(4) + $\frac{1}{2}$(2)(64) = 8 + 64 = 72 J.

The key testable concept is the work-energy theorem ($W_{net}$ = $\Delta$-KE) including all forces, and the distinction between string and rod in vertical circular motion (minimum speed at top: $\sqrt{gR}$ for string, zero for rod).