$W = \int_0^{x_0} \alpha x^n\,dx = \frac{\alpha x_0^{n+1}}{n+1}$
Given $W = \frac{\alpha x_0^{3/2}}{3/2}$: compare → $n+1 = 3/2 \Rightarrow n = \mathbf{1/2}$
Answer: $n = 1/2$
Static friction: $f_s \leq \mu_s N$ (prevents relative motion). Kinetic/sliding friction: $f_k = \mu_k N$ (constant during sliding). Rolling friction: $f_r = \mu_r N$ (very small). Typical values: $\mu_k < \mu_s$ (kinetic < static). $\mu_k$(rubber-concrete) $\approx 0.8$, $\mu_k$(ice-ice) $\approx 0.03$. Angle of friction: $\tan\phi = \mu$. Angle of repose: $\theta = \tan^{-1}\mu$ (angle at which block just starts sliding).
Newton\'s laws: 1st (inertia), 2nd ($F = ma$), 3rd (action-reaction). Momentum: $\vec{p} = m\vec{v}$; impulse: $J = F\Delta t = \Delta p$. For variable force: $J = \int F\,dt$. Conservation of momentum: in absence of net external force. Applications: rocket propulsion, collision analysis, recoil of guns.
In accelerating frame (e.g., accelerating car, rotating frame): pseudo force $F_{pseudo} = -ma$ acts on every object (where $a$ = acceleration of frame). Example: person in accelerating car feels pushed backward. In rotating frame: centrifugal force = $m\omega^2 r$ (outward), Coriolis force = $-2m\vec{\omega}\times\vec{v}$ (causes deflection, responsible for Coriolis effect in Earth\'s atmosphere → trade winds, cyclones).
Connected bodies constrained by strings/pulleys. String: massless inextensible string maintains constant length → constraint: $a_1 = a_2$ (or some relation). Atwood machine: two masses over pulley: $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$. Inclined plane with pulley: analyse forces along incline and vertical.
At rest, the microscopic contact points between two surfaces have time to settle and form stronger adhesive bonds. Once sliding begins those bonds are continuously broken and reformed imperfectly, so less force is needed to maintain motion than to start it. This is why $\mu_s > \mu_k$ for almost every pair of surfaces, and why a heavy box suddenly lurches forward the moment it starts moving — the resisting force has just dropped.
Place a block on a plane and slowly increase the tilt. At the angle of repose $\theta$ the block is on the verge of sliding, and resolving forces gives $\mu_s = \tan\theta$. This is numerically equal to the angle of friction, the angle between the normal reaction and the resultant contact force. The result is independent of the block's mass, which is why the angle at which sand or grain naturally piles up is a property of the material alone.
Rolling friction is far smaller than sliding friction because a rolling body does not drag across the surface — contact points are laid down and lifted rather than scraped. What resistance remains comes mainly from deformation of the surface and the wheel. The coefficient of rolling friction is typically a hundredth of the sliding value, which is the entire reason wheels, ball bearings and roller bearings exist. It also explains a practical detail: an underinflated tyre deforms more, raising rolling resistance and fuel consumption, while a hard wheel on a hard surface rolls most easily of all.
Assuming friction always equals $\mu N$. That is the maximum static friction. Below it, static friction equals whatever force is applied, no more — a stationary block pushed with 3 N experiences exactly 3 N of friction even if $\mu_s N$ is 10 N.
Thinking friction depends on contact area. For rigid bodies it does not. Doubling the contact area halves the pressure at each point, leaving the total friction unchanged.