Translational motion, forces, work, energy, and equilibrium in living systems

Distinguishing the pair is the entry fee for the whole section. Speed is the magnitude of the velocity vector; distance is the path length, while displacement is the straight-line change in position. A runner who finishes a lap has covered distance but has zero displacement. Resolve any vector into perpendicular components (Aₓ = A·cosθ, A_y = A·sinθ) — the single most-used move in mechanics, because the components act independently.

Pick the equation by what's missing: no displacement → use v = v₀ + at; no final velocity → use Δx = v₀t + ½at²; no time → use v² = v₀² + 2aΔx. Treat g as 10 m/s² for fast estimation (the real 9.8 rarely changes the answer choice). For a dropped object, v² = 2gh gives impact speed and h = ½gt² gives fall time — both worth knowing cold.

Decompose the launch velocity into vₓ = v·cosθ (constant) and v_y = v·sinθ (changes under gravity). At the top of the arc, v_y = 0 but vₓ is unchanged — the object is still moving. Time aloft is set entirely by the vertical problem; horizontal range is then vₓ × t. A horizontally launched object and a dropped object hit the ground at the same time from the same height, because their vertical motions are identical.

A large share of Chem/Phys points are graph-reading, not algebra. Going up the chain you take slopes (x–t slope is velocity, v–t slope is acceleration); going down you take areas (area under a v–t curve is displacement, area under an a–t curve is Δvelocity). The same logic generalizes everywhere: the slope of any y–t graph is a rate and the area under it is an accumulation — area under a force–displacement graph is work, and area under a pressure–volume graph is work done by a gas.

The third law is the most misread: the action–reaction pair acts on two different bodies, so the forces never cancel each other (they'd have to be on the same body to cancel). When you push on a wall, the wall pushes back on you with equal force — that reaction is what a swimmer or a walking person uses to move forward. The second law uses net force; a body moving at constant velocity has F_net = 0 even though many forces act.

Static friction is a range, not a fixed value — it matches the applied force exactly until that force exceeds μ_s N, at which point the object breaks free and kinetic friction (a fixed μ_k N) takes over. Both are proportional to the normal force, not to contact area or speed. On an incline, the component of gravity along the surface is mg·sinθ and the normal force is mg·cosθ, so an object stays put until tanθ > μ_s.

The lever arm is what makes torque non-obvious: a force applied through the pivot (θ = 0) produces no torque, and the same force produces maximum torque when applied perpendicular to the lever (θ = 90°). Because Στ = 0 holds about any point in equilibrium, you can choose the pivot that eliminates an unknown (place it where an unknown force acts, and that force's torque is zero) — the single most useful problem-solving trick in statics.

Levers are classified by what sits in the middle: first-class (fulcrum between effort and load, like a seesaw or the head on the neck), second-class (load in the middle, like standing on tiptoe), third-class (effort in the middle — the biceps lifting the forearm, the most common in the body). Mechanical advantage is the ratio of load to effort, equal to the ratio of the lever arms. A third-class lever's effort arm is shorter than its load arm, so its mechanical advantage is less than one — anatomically efficient because it lets a small muscle contraction produce a large, fast movement of the hand or foot. Other simple machines — pulleys, inclined planes, gears — work the same way, multiplying either force or distance while their product (work) is conserved; real machines lose some output to friction, so their efficiency (useful work out ÷ work in) is always below 100%.

Only the component of force along the displacement does work. Three consequences AAMC loves: a force perpendicular to motion does no work (centripetal force on an orbit; the normal force on a sliding block; carrying a box horizontally); friction does negative work (removing KE as heat); and lifting at constant velocity, the work you do against gravity equals mgh regardless of path. Work done by a conservative force equals the negative change in its potential energy.

The v² dependence of kinetic energy is high-yield: a car at twice the speed has four times the kinetic energy (and roughly four times the stopping distance). Gravitational PE depends only on height change, not path. Spring PE follows Hooke's law force F = -kx, giving stored energy ½kx². Energy converts between these forms — a pendulum trades PE for KE and back; a dropped ball converts mgh entirely into ½mv² at the bottom (giving v = √(2gh)).

Conservation of mechanical energy turns many two-step kinematics problems into one line: set KE_i + PE_i = KE_f + PE_f. A block sliding down a frictionless ramp of height h reaches v = √(2gh) regardless of the ramp's shape or angle, because only the height change matters. Add friction and the lost energy is f·d (friction force times path length), which appears as heat — now ΔME = -f·d. Energy is always conserved overall; "mechanical energy is not conserved" just means it left the KE+PE ledger.

Impulse is the area under a force–time curve and equals the change in momentum. For a fixed momentum change, force and contact time trade off inversely: airbags, padded dashboards, catching a ball by drawing your hand back, and a boxer "rolling with" a punch all increase Δt to decrease the force felt. This is the same area-under-the-curve reasoning as reading motion graphs.

Momentum is conserved in every isolated collision — that's what you use to solve for a final velocity. What distinguishes the types is kinetic energy: conserved in elastic collisions (billiard-ball-like), partly lost to heat/deformation in inelastic ones. In a perfectly inelastic collision the masses couple and share one final velocity (m₁v₁ + m₂v₂ = (m₁+m₂)v_f), the easiest case to compute.

The centripetal force is not a new kind of force — it's whatever real force points inward (gravity on the Moon, tension on a whirled ball, friction on a car rounding a curve). Because the force is perpendicular to the velocity it does no work, so the speed stays constant — it only turns the velocity. There is no outward "centrifugal" force; the outward feeling is inertia (your body continuing straight) in a rotating frame.

Universal gravitation sets the force between any two masses; near Earth's surface it reduces to F = mg with g = GM/r². For a circular orbit, gravity is the centripetal force (GMm/r² = mv²/r), giving orbital speed v = √(GM/r) — independent of the orbiting object's mass. The inverse-square parallel to the electrostatic force is a favorite MCAT analogy.