0
$\begingroup$

According to my textbook. The mechanical power output of a car is constant. My reasoning tells me otherwise though.

According to a lot of pages on this forum related to the driving force of a car. Some of them mention that the force of an engine varies with the speed of the car. Backed up by the fact that as the speed of the car increases, the air drag increases greatly as it varies with the the square of the speed of the car.

This therefore implies that the greater the speed the car, the greater the force of the engine (i already know about static friction as a factor contributing to the drag of the car). If power is taken to be constant, an increase in the force of the engine would lower the velocity of the car.

But shouldn't the velocity increase (as my intuition proposes i.e: mechanical power output increases with speed) when force increases according to the formular for calculating drag? Help i'm confused.

$\endgroup$
  • $\begingroup$ I don't know what "the mechanical power output of a car" even means, so I have no idea whether it is "constant" or not. But common sense would suggest either your textbook is nonsense, or (more likely IMO) you have misunderstood what it really says. For example what is the "mechanical output power" of a car which is not moving, not in gear, but with the engine running at its red-line speed? (I have no idea what your textbook thinks is the answer to that question!) $\endgroup$ – alephzero Oct 20 '18 at 11:17
  • $\begingroup$ The product $FV$ should be constant for power to be constant. $\endgroup$ – Mechanic7 Oct 20 '18 at 11:20
  • $\begingroup$ The mechanical power output is the energy used per unit time in generating the force required to drive a car i suppose/ the useful energy from the thermal energy generated by burning gasoline per unit time. $\endgroup$ – Energy Oct 20 '18 at 11:43
  • $\begingroup$ According to which textbook? And what exactly do they say, in context? $\endgroup$ – probably_someone Oct 20 '18 at 11:46
  • $\begingroup$ @Mechanic7: using the formula you are stating implies that force is inversly proportional to velocity which contradicts the fact that when the velocity of a car increases asy by a factor of 2 the drag for increases by a factor of 4. Why? Because in order to maintain a high steady velocity the force of the engine would have to counteract that of the drag force implying that the force of the engine is directly proportuonat to the velocity. $\endgroup$ – Energy Oct 20 '18 at 11:53
2
$\begingroup$

According to a lot of pages on this forum related to the driving force of a car. Some of them mention that the force of an engine varies with the speed of the car.

Force is not a property of the engine. The engine exerts a torque on the wheels. The force that accelerates a car is a force from the road on the wheels, which is a reaction to the engine's torque. In theory, you can have any gear ratio you want in a car, so the limiting factor should never be the amount of torque the engine is capable of. The reality may be different (e.g., if you don't want to shift, or because you only have 5 gears available), but this is still at least a roughly correct way of thinking about it. The reason the force of the road on the wheels is limited is fundamentally an issue of traction, not the engine.

Backed up by the fact that as the speed of the car increases, the air drag increases greatly as it varies with the the square of the speed of the car.

You need to distinguish between the force that is required and the force that is available.

The equation $P=\textbf{F}\cdot\textbf{v}$ is an identity. Let $\textbf{F}$ be the vector sum of the external forces acting on the car. These external forces are the static frictional force from the road, rolling resistance, air friction, the road's normal force on the tires, and gravity. (On a flat road, the final two forces can be neglected, because they cancel as a pair.) Then by the work-kinetic energy theorem, $P$ is the rate at which the car's kinetic energy is changing.

At low speeds, say when you accelerate suddenly from a light that has turned green, $v$ is small, so $P$ is small. The engine has more than enough power available, and you will be using a fraction of its possible power output. At these speeds, the limit on acceleration is how strong static friction is. In the usual approximations used for static friction (which are not a great approximation for rubber on asphalt, but good enough for a rough description), the maximum acceleration on a flat road is $\mu_s g$. If you try to exceed this acceleration, you burn rubber.

At high speeds, $v$ is big enough that the power required to accelerate the car may be more than the engine is capable of putting out. This is a different regime, with different limits being imposed.

The power output of the engine is certainly not constant in the sense that it never varies with time. It does vary with time, and you can easily sense this from engine sounds and how responsive the car feels. On a downhill grade, the power can be negative, which is why a hybrid can recharge its battery as you go downhill. However, an engine does have a certain maximum power output. Cars are marketed based on this figure (which is measured under optimal conditions).

$\endgroup$
  • $\begingroup$ I understand now and it's all thanks to your contributions. $\endgroup$ – Energy Oct 20 '18 at 15:14

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.