Kinetic energy

K = ½mv² uses the velocity only through v² = v⋅v = vₓ² + vᵧ² + v_z². Direction therefore costs nothing. A 0.40 kg ball whirled in a horizontal circle at a steady 3.0 m/s has K = ½(0.40 kg)(3.0 m/s)² = 1.8 J at every point, although its velocity turns continuously. Signs vanish as well: −5 m/s and +5 m/s give the same K, so kinetic energy is never negative and has no direction to draw. Components still help, provided you add their squares. A 0.50 kg ball launched at 20 m/s, 30° above the horizontal, has vₓ = 17.3 m/s and vᵧ = 10.0 m/s, so K = ½mvₓ² + ½mvᵧ² = 75 J + 25 J = 100 J. Without drag vₓ never changes, so at the top of the flight, where vᵧ = 0, only the 25 J vertical share has gone and 75 J remains. Those shares are not vector components of anything: they add as plain numbers, with no angle between them.

The formula is the output of a calculation, not a convention. Let a net force F act along a particle's line of motion. Over a small step dx it does work F dx, and Newton's second law rewrites that as m(dv/dt) dx = m v dv, because dx/dt = v. Adding the steps from rest up to speed v gives ∫₀ᵛ m v dv = ½mv². So K is the work that must be done on a body to bring it from rest to v, and the work it can do on whatever brings it back to rest. The square has a physical reading. Under a constant stopping force, a body twice as fast takes twice as long to stop and moves twice as fast meanwhile, so it slides four times as far and the force does four times the work. A 1500 kg car at 16 m/s holds ½(1500 kg)(16 m/s)² = 1.92 × 10⁵ J, and a steady 8.0 kN braking force must act over (1.92 × 10⁵ J)/(8.0 × 10³ N) = 24 m to remove it. At 32 m/s the same force needs 96 m.

Because K is quadratic in speed, a change in K depends on where on the speed scale it happens, not only on its size. Factor the difference: ΔK = ½m(v₂² − v₁²) = ½m(v₂ + v₁)(v₂ − v₁), so a given Δv costs energy in proportion to the average of the two speeds. A 1200 kg car needs 6.0 × 10⁴ J to go from 0 to 10 m/s, but 1.8 × 10⁵ J to go from 10 to 20 m/s: three times as much for the same 10 m/s. Run it backwards and braking from 20 m/s to 10 m/s already sends three quarters of the car's kinetic energy into the brakes. The same arithmetic governs stopping under a fixed braking force, where the distance goes as v²: a car at 60 km/h needs (60/50)² = 1.44 times the distance it needs at 50 km/h, 44% more road for 20% more speed. When a question asks what it costs to add speed, the starting speed is part of the data.

Speed depends on who measures it, so kinetic energy does too. Seen from a frame moving at constant velocity u, velocities obey v = v′ + u, and squaring gives K = K′ + m u⋅v′ + ½mu². A flight attendant pushes a 25 kg service cart from rest to 1.0 m/s towards the nose of an airliner cruising at 230 m/s. In the cabin frame ΔK′ = ½(25 kg)(1.0 m/s)² = 12.5 J. From the ground the cart goes from 230 to 231 m/s, so ΔK = ½(25 kg)[(231 m/s)² − (230 m/s)²] ≈ 5.76 × 10³ J. Neither number is wrong. Work depends on the frame too, because displacement does: during a one-second push the attendant's hand moves some 230 m over the ground, and W = ΔK holds in each frame separately. The gap is ΔK − ΔK′ = u⋅Δp = (230 m/s)(25 kg m/s) = 5750 J. It vanishes whenever the momentum is unchanged, which is why every inertial frame agrees on the kinetic energy a collision dissipates.

For a system of particles the kinetic energy is K_sys = Σ ½mᵢvᵢ², a sum of terms that are each positive, so nothing in it cancels. It is a state quantity: it depends only on the velocities at one instant, so ΔK between two states is fixed by the states themselves, however the motion in between unfolded. Work is the other kind of quantity, a transfer that happens during a process. Measuring each velocity from the centre-of-mass frame, which moves at V_cm = Σmᵢvᵢ/M, splits the sum cleanly into K_sys = ½MV_cm² + Kᵢₙₜ, because the cross term contains Σmᵢvᵢ′, which is zero in that frame. A 2.0 kg rifle fires a 10 g bullet at 400 m/s and recoils at 2.0 m/s. Both start at rest and their momenta are equal and opposite, 4.0 kg m/s each, so V_cm stays zero. Yet K = p²/(2m) gives the bullet 800 J and the rifle only 4.0 J: 804 J in total, all of it internal, which ½MV_cm² would report as nothing.

Two assumptions sit inside ½mv². The first is that every part of the body shares one velocity. A hoop rolling without slipping at 5.0 m/s has its top moving at 10 m/s and its bottom momentarily at rest, and its total kinetic energy is twice ½MV_cm²; the extra is rotational, which the course later writes as ½Iω². A flexing diver or a vibrating spring hides internal energy in the same way. The second assumption is low speed. Relativity replaces ½mv² with (γ − 1)mc², where γ = 1/√(1 − v²/c²). Expanding for small v/c gives ½mv² + (3/8)mv⁴/c² + …, so the classical value runs low by roughly (3/4)(v/c)²: 0.75% at 0.1c and 19% at 0.5c. For anything a mechanics course launches the correction is invisible: a satellite at 7.7 km/s has v/c ≈ 2.6 × 10⁻⁵, and ½mv² is good to better than one part in a billion.