Energy with external and dissipative transfers

An energy equation is a statement about a system, so the first line of a solution names what is inside. Then sort every interaction. A conservative pair with both members inside, crate and Earth or block and spring, appears as a potential energy change ΔU. Kinetic friction between two surfaces that are both inside appears as a thermal energy change ΔEₜₕ. A force from anything outside, a hand, a rope, a cable from a motor, appears as work Wₑₓₜ, positive when energy flows in and negative when it flows out. What remains is ΔK + ΔU + ΔEₜₕ = Wₑₓₜ. For a crate winched up a rough ramp, take crate + ramp + Earth: gravity becomes U, the rubbing becomes Eₜₕ, and the only transfer across the boundary is the cable's pull.

Let a block slide across a fixed floor and slow under kinetic friction fₖ. Newton's second law for the block, m v dv/dx = −fₖ, integrates over the sliding distance d to ΔK = −fₖ d. Now take block + floor as the system: nothing outside does work and U is unchanged, so ΔEₜₕ = −ΔK = fₖ d. That energy is shared between the two surfaces, which is why the rubbing pair belongs inside the boundary. Two details fix the number. Use the distance slid along the path: a 4.0 kg crate pushed 3.0 m across a floor with μₖ = 0.25 and back again produces 9.81 N × 6.0 m = 58.9 J of thermal energy, with zero displacement. And use sliding relative to the other surface: a parcel on a moving belt is charged for the distance it slips along the belt, not the distance it covers over the ground.

Lower a 25 kg box 1.8 m at steady speed on a rope. With Earth inside the system, ΔK = 0 and ΔU = −mgh = −25 × 9.81 × 1.8 = −441 J, nothing slides, so Wᵣₒₚₑ = −441 J. With Earth outside, the box alone stores no potential energy; gravity does external work W_grav = +441 J, and ΔK = W_grav + Wᵣₒₚₑ = 0 gives Wᵣₒₚₑ = −441 J again. The two boundaries tell one story. Choose the one that shows what you need: to find what the rope takes out, either works; to ask where the energy goes, keep Earth inside so the drop in U is on the page. The negative sign of Wᵣₒₚₑ is not an error. It says energy leaves the system through the rope, ending up in the hands, a brake, or a winch motor running as a generator.

Wₑₓₜ = ∫F⋅dr follows the point where the outside force is applied, and for anything that is not a single particle that point need not move with the body. A 55 kg skater pushes off a rink wall and glides away at 2.0 m s⁻¹. The wall's push is large, but the palm on the wall stays put while they touch, so the wall does zero work. Yet the skater gains ½ × 55 kg × (2.0 m s⁻¹)² = 110 J. With the skater as the system, ΔK + ΔEₜₕ + ΔEᵢₙₜ = 0, so the skater's internal chemical energy fell by at least 110 J, and by more once the warmth of working muscles is counted. A car is the same: the road's static friction pushes it forward but does no work, because each patch of tread touching the road is momentarily at rest. The fuel pays. This is where the three-term ledger stops and an internal energy term has to be added.

Divide the energy equation by dt and it becomes a statement about rates: Pₑₓₜ = dK/dt + dU/dt + dEₜₕ/dt. Each outside force contributes F⋅v, and kinetic friction heats at fₖ vᵣₑₗ, the power loss. A 1500 kg car holds 20 m s⁻¹ on its brakes down a road that drops 5.0 m for every 100 m travelled, so sin θ = 0.050. Take car + Earth as the system and ignore drag and rolling resistance. ΔK = 0 and nothing outside does work, so the potential energy lost every second, mgv sin θ = 1500 × 9.81 × 20 × 0.050 = 14.7 kW, is all converted by pads rubbing on discs, the output of about five 3 kW kettles. Brakes fade on a long descent because that thermal energy arrives faster than the discs can pass heat to the air, and heat leaving across the boundary is a transfer this mechanical ledger never listed.

Haul a crate at steady speed up a ramp of angle θ to a dock of height h. The ramp is L = h/sin θ long, the cable pulls with T = mg(sin θ + μₖ cos θ), and Wₑₓₜ = T L = mgh(1 + μₖ cot θ). Only mgh is stored, so η = mgh/Wₑₓₜ = 1/(1 + μₖ cot θ), independent of mass and speed. With μₖ = 0.30, a 20° ramp gives η = 1/(1 + 0.30 × 2.747) = 54.8% and a 40° ramp gives 73.7%. The shallow ramp asks for less force, but it drags the crate along a longer rubbing path, and friction charges by the metre. Efficiencies in series multiply: an 85% efficient winch motor on the 20° ramp stores 0.85 × 0.548 = 46.6% of the electrical energy it draws, and the other 53.4% becomes thermal energy, part in the motor windings and part in the ramp.