What is the formula of the exerted force

VI.1.1 Definition of work

VI.1.2 Extension to variable forces and any way

VI.1.3 Acceleration work

VI.1.4 Lifting work

VI.1.5 Tensioning work

The concept of work is already known from everyday language usage. If you want to move masses, you have to do work for it. To put it more precisely, in physics this means that when masses move, forces have to do work.

A heavy body is attached to a rope and pulled along a straight path. In order to have a measure of the force, you can, for example, attach a spring balance to one end of the rope. Now you can pull the rope at different angles to the earth's surface in order to move the body. As borderline cases, it should be drawn parallel to the ground and perpendicular to it. In this two-dimensional case, the rope indicates the direction of the acting force.

Now the body is to be pulled a fixed distance s with constant force. The experiment shows that you need the least force at an angle of 0 , i.e. drawn parallel to the earth. The larger the angle, the more force the spring balance shows. At an angle of 90 you only lift the body up and not move it in the s direction at all. It turns out that only the force component along the path is effective for the movement of the body.

The product of effective force and path is called work.

Unit consideration:

[F s] = [F] [s] = 1N 1m = Nm. The composite unit Nm is denoted by Joule, as the unit symbol J. From the definition of Newton it also follows: 1J = 1 kg m2 s-2. In the earlier common csg system, the unit of work is denoted by erg, where 1 erg = 10-7 J is.

The formula for work given in definition IV.I is a scalar product. It is known from geometry that the product of two scalars is equal to the content of the rectangle that can be spanned with the side lengths of the two scalars. Drawn in the coordinate system, this means that the area under the force-displacement diagram is a measure of the work done. (With force here, of course, is meant the force along the way.)

Let us consider this working diagram with a variable force along a straight path. Since force here only refers to the force component along the path, both the amount of force and its direction can now vary. In this case, the work can be understood as the area under the curve of the force. Such an area is calculated with an integral.

It then applies

or for any straight line that does not have to start at the zero point: .

Let us now assume that the line cannot run along a straight line, but arbitrarily in space. Then definition IV.1 only applies to infinitely small path elements in which the route can be assumed to be straight. The work has to be summed up over these small path elements.

The following applies to small path elements,

with the definition of the vector from Chapter I. .

Summed up over all sections follows:

The work WFROM along a section from point A to point B corresponds to the part under the curve of the force component F.S., which is constructed by the perpendicular to the two boundary points.

When deriving the concept of work, we neglected the speed of the movement, so we considered a quasi-stationary movement. The definition of the force as the origin of every acceleration suggests that we should examine this neglect more closely. I can also pull a box so that it gets as fast as possible. What then follows from the definition of motion?

In this case, the force that causes a movement serves to accelerate the body of mass m. With Newton's second axiom, such a force can always be written as . Let's put this definition of force in the formula


then follows:



Now we are writing






The acceleration work depends only on the mass m of the accelerated body and on the initial and final speed. The direction and the path in which the acceleration work takes place is also irrelevant for the value of the acceleration work performed.

From and follows:

A more precise interpretation of the acceleration work is given in Chapter IV.3.

Another form of work that can be done is lifting bodies against the force of weight. In a thought experiment, let a body of mass m take an arbitrary path from a height h1 to a greater height h2 upscale. Let h be defined as positive for an upward movement. This process should now be carried out as slowly as possible, i.e. without acceleration.

The external force that has to be applied is the counterforce to the weight acting downwards on the body:


The following applies to work

with definition IV.2 .

The work is only done with the part of the force that points in the direction of movement, i.e. only with component Fa cos.

It follows.

With applies

with G = - mg

The lifting work is therefore also independent of the path on which the height is reached. It is only dependent on the mass of the lifted body and the absolute height difference h.

The work that has to be done against the weight depends only on the endpoints. This is a property of weight and some other forces. These forces are taken together and with the term conservative forces designated. From the consideration that the work from point A to point B depends only on the height difference h, and also from point B to A only on the height difference - h, it follows directly that the work goes back from any point A via any path must be zero after point A. The integration over a closed path is called a ring integral.

It is known from experience that the force required to deflect a spring increases constantly with increasing deflection. So is , or F x. The constant of proportionality depends, for example, on the material of the spring or the number of turns. It is therefore a material constant and is denoted by k or D in different ways.

This empirical result is formulated in Hook's law:

Hook's force is also a conservative force.

The work can then be done with

calculate as

with const = 0

In particular, the following applies to the deflection x from the rest position