Newtonian Mechanics - Force Required To Drive Car

Diesel powered cars typically get better gas mileage or fuel efficiency than gasoline powered vehicles. Also, some drivers feel that they get a better value Many car manufacturers are designing vehicles that can be powered by ethanol because it is a cost effective fuel made from renewable resources...There are two components to the force required to drive a car along. aerodynamic resistance and rolling resistance. Aero is by far the larger of the Power approximately depends on the cube of the speed. Double the speed you need about 8 times the power. But it takes very little power to move a...Hydrogen fuel cell cars are powered by an electric motor and are therefore classified as e-cars. The common abbreviation is FCEV, short for "Fuel Cell That depends on the conditions under which the hydrogen for the fuel cell vehicles was produced. Hydrogen production requires electrical energy.The required engine power for a car driving on a flat surface with constant speed 90 km/h with an aerodynamic resistance force 250 N and rolling resistance force 400 N and overall efficiency 0.85 - can be Engineering ToolBox, (2011). Car - Required Power and Torque. [online] Available at: https...Average Power Exerted By a Speeding Car Up an Inclined Plane - Work & Energy Physics Problems. Thermodynamics - Power Needs of a Car to Climb a Hill.

What power is required from the car's engine to drive the... - Quora

A. A 710kg car drives at a constant speed of 23m/s . It is subject to a drag force of 500 N. B.What power is required from the 6 years ago. power is work per unit time which comes out to force times velocity at constant velocity. a 2deg slope means gravity has little effect so the power needed in this...(b) Let P is the power is required from the car's engine to drive the car on up a hill with a slope of 2 degrees. At this slope, force will beElectric cars, just like gas powered cars, come in a wide range of sizes, and have motors with a wide range of power. Cars motors are often rated in the The current coils of my schematic are derived from the starter coils of a single phase induction motor. These voltage coils, VC1 and VC2, will exhibit...Clearly if the car was travelling at 36kph (10m/s) whilst the motor was producing this power/torque/speed then the gearing would need to The required vehicle torque will depend on the roughness of the terrain and the maximum angle of the road you wish to travel. I suggest you edit your...

Hydrogen fuel cell cars: what you need to know | BMW.com

The chassis includes a power train (power transmission), a running gear, steering and braking systems as well. The power train carries the power from the engine to the car wheels.Ignoring air friction, when does the car require the greatest power? (a) When the car first accelerates from rest, (b) just as the car reaches its maximum speed, (c) when the car reaches half its maximum speed. (d) The question is misleading because the power required is constant. (e) More information...Most modern cars require the battery to be present while the engine is running, otherwise it will just die. Of course if the alternator dies, then the car starts drawing power from the battery to keep going. That only will last for a short time though.The battery is also required to power the electromagnets in an alternator, so that it may generate power. All lighting and accessories depend on On a modern car no. Older completely, mechanical vehicles will run without a battery after starting because everything is driven from the motor or...Power Inverters are little boxes that plug into your cigarette lighter and handle the conversion. Most modern electronic, solid-state devices don't require much. If you're just idling at, say, 1000 rpm, and the only load on the car is from the draw from the inverter, an average car should be able to run...

Christina, sorry to butt in here, however I tried to answer your escalator question, deleted it due to a mistake, then couldn't re-answer it. So here it is:

Work Done = Force x Displacement.

As it's only the DISPLACEMENT you undergo, due to the FORCE acting on you, the length of the escalator does not subject in this case.

Assuming you commute at consistent pace (or velocity on this case, should you like), all the forces acting on you must be in steadiness. Here, those forces are gravity pulling you down, and the escalator pushing up against you, preventing you from accelerating to the ground.

So the drive the escalator pushes you up with shall be equivalent in magnitude (size) - however opposite in course - to that of gravity pulling you down, i.e. your weight mg, with m=62kg, and g=10ms^2.

As you might be coping with vectors right here, you most effective take the elements of the forces which act parallel to the course of your displacement. So if your pressure acted at 30 levels to the ground, however you travelled instantly up, you wouldn't take the complete magnitude of the drive, but break up it into 2 elements: 1 parallel to the ground, and the different perpendicular to the ground, and use the latter element to your sums.

BUT on this case, both your displacement and the 2 forces acting on you are all vertical, making your sums a little easier.

So you'll just multiply your weight mg by way of your displacement (4.5m), and get your Work Done.

BOTH escalators will do the SAME amount of Work on you, as they both act directly up towards you, to prevent you from accelerating down beneath gravity. This is primarily based on the assumption that you're transferring at constant speed (velocity) on each escalators, as your question does not indicate in a different way. Similarly, as there is no point out of the frictional forces or air-resistance which would resist the escalator moving you in the horizontal component of your displacement, you don't want to consider horizontal forces, or subsequently, the duration of the escalator.

The query is tricking you via including the length of the escalators, and asking about each UP and DOWN escalators, to see if you already know your stuff.

Now you should!