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From the work energy-theorem, we know that: $$ W_{net}=KE_{final}-KE_{0} $$ Now, imagine that there is a car that is initially at rest. Eventially, at some point in time, it attains kinetic energy of 1000$J.$ This means that the work done on the car is also 1000$J.$ But we know from the second law of thermodynamics that as energy is converted between different forms, it always becomes less useful. Is the intuition here, that the work done on the car is only useful chemical energy? In other words, as chemical energy is transformed into say heat+kinetic energy, the work represents precisely the amount of chemical energy useful in being transformed to the kinetic energy?

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Broadly, yes. Let's suppose the car has a petrol engine and is initially at rest. The engine will perform work which causes the car to accelerate and acquire kinetic energy. That process is not perfectly efficient, however, as some of the energy generated by the engine will be given off as heat and noise, and some will be given off as heat and noise resulting from friction at various points within the drive train (eg between the tyres and the road), and some of the energy will be used up in overcoming air-resistance, and some will be used up in ancillary processes (eg powering the dashboard instruments and driving the windscreen wipers if it is raining). So the kinetic energy gained by the car will generally be considerably less than the overall energy consumed by the engine.

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    $\begingroup$ Got it. So then the work done by the engine is a lot more than the work represented by the work-energy theorem then. $\endgroup$ Commented Oct 21, 2022 at 13:59
  • $\begingroup$ Indeed, yes you've got that exactly right. $\endgroup$ Commented Oct 21, 2022 at 14:23
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Mechanical work measures how much kinetic energy is transferred to or subtracted from an object. So $ W = \pm \Delta K $. When object is loosing kinetic energy, then work done is negative. Consider a car running at some speed and turning fully breaks-on until it stops. In this case road kinetic friction has done negative work on a car, in which all car's kinetic energy was converted into heat between contacting surfaces - tires & asphalt. In this negative work scenario chemical energy does not participate at all. All in all - work done may be variously related with type(s) of energy exchanges. Main point is decrease or increase of KE.

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Following on from some of the other answers and having a bearing on the question in that a car's efficiency depends on the power source and actually moving the car against dissipative forces.

Modern gasoline engines have a maximum thermal efficiency of more than 50%, but road legal cars are only about 20% to 35% when used to power a car. In other words, even when the engine is operating at its point of maximum thermal efficiency, of the total heat energy released by the gasoline consumed, about 65-80% of total power is emitted as heat without being turned into useful work, i.e. turning the crankshaft. Approximately half of this rejected heat is carried away by the exhaust gases, and half passes through the cylinder walls or cylinder head into the engine cooling system, and is passed to the atmosphere via the cooling system radiator. Some of the work generated is also lost as friction, noise, air turbulence, and work used to turn engine equipment and appliances such as water and oil pumps and the electrical generator, leaving only about 20-35% of the energy released by the fuel consumed available to move the vehicle.
Wikipedia - energy efficiency

An electric vehicle's electric motor converts more than 90% of the electrical energy it consumes into useful work, so its efficiency is more than 90%. Add to that the fact that the electric motor can recover some of the energy it puts into moving the car by generating electricity as it slows, and the efficiency improves still further.
An Imperial gallon (4.5 litres) of petrol contains the equivalent energy of just over 40kWh of electricity. That means a typical, middle-of-the-road electric vehicle with a 64kWh battery (useable charge) is carrying roughly the same amount of energy as 1.6 gallons (7.2 litres) of petrol. A 1.5-litre petrol car of about the same size as that electric vehicle will optimistically manage 55 miles/gallon (23 km/litre), so 1.6 gallons (7.3 litres) of petrol will take it 88 miles (142 km). At a realistic 3.7 miles (6 km/kWh), the 64kWh electric vehicle will travel almost 235 miles (378 km). The difference of 148 miles (238 km) is due to the efficiency of the electric power train and the energy recovered by regenerative braking.
Autocar - How efficient are electric cars?

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  • $\begingroup$ if you talk about efficiency this way then a truly fair comparison should also include the energy costs of making electricity and its distribution to charge the batteries and their digging minerals from the earth to burying them versus oil drilling/refining distribution, etc., never mind the very difficult comparison of environmental costs. $\endgroup$
    – hyportnex
    Commented Oct 21, 2022 at 14:55

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