Work and Force x Distance Relation

Question

I'm trying to understand the work-energy theorem, and I understand the relation here:

$$W = \int \vec F \cdot d\vec{x}$$

But its the process that I need help clarifying a hypothetical.

Work is equal to the force applied to an object over a distance x. Let's say that object 'one' is in a vacuum with 0 velocity before a constant force from object 'two' is applied to the first object over 10 metres by a push. If the force is to be constant over time (the time it was pushing object two over those 10 metres), would that mean that object 'two' must be also be increasing in velocity (from some third force) to maintain it's constant force on object 'one' over the 10 metres? Since as soon the the force from object 'two' is applied to object 'one', the velocity of object 'one' increases. And so a constant force to object one requires a third force to be acting on object two, and so on with object four to object three. Is this

Answer 1

The OP's worry is certainly true, if the force exerted on ONE by TWO is a contact force.

But, there are some non-contact forces. These forces may settle your worry.

First example, the expelling jet force: Saying that object TWO is the combustion gas, expelled from inside the object TWO. We may adjust the gas expelling rate to obtain a constant acceleration.

Second example, the object TWO is a set of parallel electric plates carrying opposite charges. This creates a constant electric field between these plates. The object ONE is a charged particle site in between the two plates. This arrange render a constant force upon the object ONE.

Answer 2

It is certainly true that many examples that one is given considers just one body being accelerated by a constant force and no mention is made as to the origin of the force. For example a train accelerating on a railway under the influence of a constant (contact) force on the train due to the track (and Earth), a ball in free fall under the influence of the constant non-contact gravitational attraction due to the Earth.

It is important to first define the system you are considering and that in turn will enable one differentiate between the internal forces (Newton third law pairs within the system) and the external forces.

The way you have asked your question the system has to include the body $2$ which is producing the force on the object under consideration, body $1$.
In the two examples I have given the systems are; train (body $1$) + track/Earth (body $2$) and ball (body $1$) + Earth (body $2$).
In each case the two internal forces (force on body$1$ due to body $2$ and force on body $2$ due to body $1$) are equal in magnitude and opposite in direction (Newton's third law).
Since each body has a force on it, each of the bodies must accelerate however in examples of this type is is assumed (although not always stated) that the mass of body $1$ is much, much less than the mass of body $2$.
This has the consequence that the acceleration, change in velocity and change in kinetic energy of body $2$ is very small compared with that of body $1$ and so can be neglected although they must exist as a consequence of momentum conservation.
So the train moves "forward" but the track is not seen to move "backwards" and the ball is seen to "fall" downwards but the Earth is not seen to move upwards.

In this sort of example one can then look at the motion of body $1$ only, ie define the system as body $1$ only and assume a constant external force (due to body $2$ acting on body $1$ and not worry as to those bodies outside the system under consideration producing the external force.

Answer 3

If the force is to be constant over time (the time it was pushing object two over those 10 metres), would that mean that object 'two' must be also be increasing in velocity (from some third force) to maintain its constant force on object 'one' over the 10 metres?

Not necessarily. Object 2 doesn't necessarily have to "move" along with object 1 while applying a contact force to object 1 over a distance $x$.

Let object 1 be the ground (the earth) and object 2 be a car. In order for the car to accelerate forward a clockwise torque is applied to the drive wheel(s). The torque applied to the tires exerts a static friction force backwards on the ground. Per Newton's 3rd law the ground exerts an equal and opposite static friction force forward on the car. Neglecting air resistance, the static friction force exerted by the earth on the car is the only external horizontal force acting on the car, and is therefore responsible for the acceleration of the car.

From Newton's 2nd law the acceleration of the car, where $F$ is the static friction force exerted by the ground, is

$$a_{car}=\frac{F}{m_{car}}$$

Since, from Newton's 3rd law, the car exerts an equal and opposite static friction force on the earth, the earth will undergo an angular acceleration $\alpha$ of

$$\alpha_{Earth}=-\frac{F}{M_{Earth}R}$$

where $R$ is the radius of the earth.

Since the mass of the earth is so much greater than the car, its acceleration would be infinitesimal (immeasurable). So although the car accelerating over a distance $x$ is in continual contact with the ground, the point of contact with the ground continually changes. The point of contact does not "move" (accelerate) with the car.

Let's take another example involving a man running behind a cart pushing it with a constant force $F$ over a distance $x$. In this case a "third force" is in fact necessary for the man to accelerate with the cart. That third force is the static friction force the ground exerts forward on the man's feet in reaction to the force the man's feet exert back on the ground.

Hope this helps.

Answer 4

Object one can be a car that won't start. While the driver press the clutch pedal, another person push the car with a constant force. It accelerates. When the velocity is big enough, the driver let the pedal free, and pray for the start.

The helping guy is object two, and it is enough.