$W = \int_0^{18^{1/3}} 3x^2\, dx = \left[x^3\right]_0^{18^{1/3}}$
$= (18^{1/3})^3 - 0 = 18^{3/3} = \mathbf{18}$ J
Answer: 18 J
Work: $W = \vec{F}\cdot\vec{d} = Fd\cos\theta$. Unit: Joule (J) = N m. Positive work: force in direction of motion. Negative work: force opposing motion. Zero work: force perpendicular to motion. Variable force: $W = \int F\, dx$ (area under F-x graph). Conservative force: work done path-independent (gravity, spring, electrostatic). Non-conservative: work done path-dependent (friction, viscosity).
Kinetic energy: $KE = \frac{1}{2}mv^2$. Gravitational PE: $U = mgh$. Spring PE: $U = \frac{1}{2}kx^2$. Conservation of mechanical energy (no friction): $KE_1 + PE_1 = KE_2 + PE_2$. Work-energy theorem: $W_{net} = \Delta KE$. Work done by non-conservative forces = $\Delta ME$ (change in mechanical energy).
Elastic collision: KE conserved; momentum conserved. For equal masses: velocities exchanged. Inelastic collision: KE not conserved; momentum conserved. Perfectly inelastic: objects stick together. For perfectly inelastic: $v_f = (m_1 v_1 + m_2 v_2)/(m_1 + m_2)$. Coefficient of restitution $e$: ratio of relative speed after to before collision. Elastic: $e = 1$. Perfectly inelastic: $e = 0$.
Power: $P = W/t = Fv\cos\theta$. Average power: $\bar{P} = W/t$. Instantaneous power: $P = F\cdot v$. Efficiency: $\eta = P_{output}/P_{input} \times 100\%$. Practical machines always have $\eta < 100\%$ due to friction and heat losses. Motor power: if a motor lifts mass $m$ through height $h$ in time $t$: $P = mgh/t$.
When the force changes along the path, $W = Fd\cos\theta$ no longer applies and the work must be integrated: $W = \int \vec{F}\cdot d\vec{s}$. Graphically this is the area under the force-displacement curve, which is often the fastest route in an exam — a triangular F-s graph gives $W = \frac{1}{2}F_{max}s$ directly. A spring is the standard case: the restoring force grows linearly as $F = kx$, so the work stored is $W = \frac{1}{2}kx^2$, and that same factor of $\frac{1}{2}$ appears because the average force over the stretch is $kx/2$, not $kx$.
A force is conservative if the work it does depends only on the start and end points, not on the path taken, and is zero around any closed loop. Gravity, the spring force and the electrostatic force qualify; friction and air resistance do not. Only conservative forces can be assigned a potential energy, which is why there is a gravitational PE and a spring PE but no "friction PE". When non-conservative forces act, mechanical energy is not conserved and the work-energy theorem must be written as $W_{conservative} + W_{non-conservative} = \Delta KE$.
Power is the rate of doing work, $P = dW/dt$, and for a constant force moving at velocity $v$ it becomes $P = \vec{F}\cdot\vec{v} = Fv\cos\theta$. Average power over an interval is total work divided by total time, while instantaneous power is the value at one moment — a car accelerating from rest has low instantaneous power at the start and much higher power later, even at constant engine output, because v is rising. This is also why a vehicle climbing a hill at constant speed needs $P = mgv\sin\theta$ on top of whatever power overcomes friction. The commercial unit, 1 horsepower, equals 746 W, and 1 kilowatt-hour is an energy of $3.6 \times 10^6$ J, not a power.
Using $Fd$ when the angle is not zero. Work is the product of force and displacement along the force. A horizontal push on a box moving horizontally gives $Fd$; carrying the same box horizontally at constant height does zero work against gravity, because the angle is 90°.
Forgetting that work can be negative. Friction and a retarding force do negative work, which removes kinetic energy. A body slowing down has negative net work done on it.