pushing an object in a car

[freefall]

Say you're driving in a car pretty fast. You hold out an object in your hand and push it out in front of you, the same direction that the car is moving.

 

You would feel no more resistance than if you were travelling at 0 m/s right?

 

But aren't you doing more work on the object? You're forcing it over a greater distance because the system is moving faster? But it wouldn't feel any different...

 

Consider you and the car to be the external system.

 

Could someone explain?

 

This is related to what someone told me: that in a car, it takes more gas to increase your speed by a constant increment when you're moving faster because the engine needs to do more work on the car. But it doesn't seem like it would be an "harder" for the engine to speed it up.

 

Thanks very much!

[[Tycho?]]

The force would indeed be the same to push and object in a car and out of a car. When you are moving at a constant velocity it means there are no net forces acting on the system. When this happens there is no way to distinguish the moving frame from the still frame (from a physics point of view). Maybe its not the car thats moving at all, maybe its the earth rotating under the wheels of the car. From this point of view the question no longer makes sense.

 

Eh, that explanation wasn't very good. Stick around for some other, better explanations.

 

Oh, and there would be zero work done in any case, if you move an object horizontally there isn't work done, the force is perpendicular to the motion (force downward of gravity, motion forward of hand).

[freefall]

there would be zero work done in any case,

right, I mean you accelerate the object.

[5614]

The kinetic energy of an object is given by

1/2mv^2

or

a half * mass * velocity squared

So the difference between 1 and 2 is the same as between 50 and 51, however because it is velocity squared the difference between 50 squared and 51 squared would be bigger than difference between 1 squared and 2 squared.

 

But now we have to start talking in terms of frames.

 

You say that the object is already moving at a speed because the car is moving at a speed and the object is in the car.

 

But relative to the car the object is not moving. This is the frame you must use in this case.

 

Relative to the car the object IS moving at 0m/s therefore it is irrelevant what speed the car is moving at.

[Jacques]

This is related to what someone told me: that in a car, it takes more gas to increase your speed by a constant increment when you're moving faster because the engine needs to do more work on the car. But it doesn't seem like it would be an "harder" for the engine to speed it up.

That's true because of air resistance. Air resistance is proportional to the square (or the cube, can't remember) of the speed. On the moon it won't be true because there is no air.

[MattC]

Just to clarify, there IS work being done, when you push something, even if you push perpendicular to the pull of gravity. When trying to determine if "work" in the technical sense is being done, seeing if it is "work" in the practical sense is useful, as they pretty much coincide. For instance, it is work to push something. ... and in the technical sense, it is also "work." Work, in physics, is, among other definitions (amounting to the same thing, I believe), the sum of kinetic and potential energy changes. When you move something, you add to it's kinetic energy. If it stops moving due to resistance after you push it, work is still done - that work simply was translated into heat in the form of friction with air.

 

In the case of a car, the simplest explanation is that the car engine works most efficiently when you are driving around 50 mph (varies from engine to engine, probably can vary a good bit; I'm no expert on cars). So if you're going under 50 (using 50 mph peak efficiency in this example; this is an assumption) and you speed up, until you hit 50, you get more distance out of each gallon of gas, until you are moving over 50. Then you get diminishing returns.

 

In your example of pushing something while in a moving car, work is done on the car and everything inside when you first start the car up (when you get moving). Once you are at your desired speed, the car needs to do *less* work to sustain that speed than to accelerate - the work that it needs to do is done to make up for the loss of energy through friction. Assuming your car spends this energy to keep the car speed constant, the inside of the car, because the windows are closed and the air is "still," is an inertial frame of it's very own. In other words, as long as those windows aren't open and air isn't rushing in, you can push or pull and the effect is no different than if you were pushing and pulling (to the same degree) while NOT in a moving car.

 

Here's another thing to think about - the earth is spinning around the sun at a very fast speed (around 67000 mph, I believe). So the earth is constantly moving. Does it take more effort to run with the earth, or to run in the other direction (this thought experiment is a little convoluted. Just assume you can run either along with the earth, or in the opposite direction that the earth is moving, without actually stepping off the earth). Further, the sun and the entier solar system is moving, presumably outward from the center of the universe, at a speed that I would guess (I have no data for the actual speed; google it if interested, I'm sure it's high) is much faster than even the 67000 mph the earth is moving around the sun. But really, why use the sun as the reference point? Or why use the center of the solar system? The answer is that you shouldn't - set a convienent inertial frame of reference for physics problems like yours, and it'll make things much cleaner.

I hope that helps a little!

[freefall]

assume no friction.

When you accelerate your car, the road forces it forward, right? The faster you're moving, the longer the distance the road has to force it to increase your speed by some constant amount. f x d = work. So, how does that not correlate to more work done on you car by the road (which means more work done by the engine)?

[5614]

So, how does that not correlate to more work done on you car by the road (which means more work done by the engine)?

What?

[Connor]

'']Oh, and there would be zero work done in any case, if you move an object horizontally there isn't work done, the force is perpendicular to the motion (force downward of gravity, motion forward of hand).

 

there need not be an external force to push against, the inertia of the object requires work to be done on it. Imagine a 1 kilogram a weight and a 50 kilogram weight in space. The 50 kilogram weight will require more work to move because of its greater inertia, despite the lack of gravity

[freefall]

okay guys you gotta help me out here.

 

It seems to me that in order for the car to increase its speed by, say, 1 m/s, it takes more work the greater the velocity of the car.

 

KE=1/2mv^2. If v is small, KE doesn't have to increase much to get v to increase by 1 m/s. But as v gets large, KE has to increase by a larger margin to get v to increase by 1 m/s. How am wrong?

 

Because MattC says it doesn't take more work.

 

Thanks very much.

[swansont]

okay guys you gotta help me out here.

 

It seems to me that in order for the car to increase its speed by' date=' say, 1 m/s, it takes more work the greater the velocity of the car.

 

KE=1/2mv^2. If v is small, KE doesn't have to increase much to get v to increase by 1 m/s. But as v gets large, KE has to increase by a larger margin to get v to increase by 1 m/s. How am wrong?

 

Because MattC says it doesn't take more work.

 

Thanks very much.[/quote']

 

 

KE is not an invariant quantity when you transfer between frames, so the change in KE going from 1—>2 mph in your own frame and e.g. 51—>52 mph in an external frame is the wrong comparison.

 

The equation for KE is not linear in v, so a given amount of work will not result in the same change in v at different speeds — it will result in the same change in KE.