If we push or pull a body meeting no hindrance to what we are doing, the result will be an acceleration of the body. The increase in kinetic energy taking place in this connection is called the work \(W\) performed by the force: \[W= \frac{mv_{2}^{2}}{2} - \frac{mv_{1}^{2}}{2}\] According to Newton’s law, the acceleration of a body and hence also the increase in its kinetic energy, is determined by the vector sum of all the forces applied to it. Therefore, in the case of many forces, the formula \(W = (mv_{2}^{2}/2) - (mv_{1}^{2}/2)\) expresses the work performed by the resultant force. Let us express the work \(W\) in terms of the magnitude of the force.

For the sake of simplicity, we shall restrict ourselves to the case when motion is possible only in one direction – we shall push (or pull) a cart of mass \(m\), standing on rails (Figure 1).

According to our general formula for uniformly accelerated motion, \(v_{2}^{2} - v_{1}^{2} = 2as\). Therefore, the work performed by all the forces over a distance \(s\) is \[W= \frac{mv_{2}^{2}}{2} - \frac{mv_{1}^{2}}{2} = mas\] The product \(ma\) is equal to the component of the total force in the direction of the motion. Consequently, \(A =f_{t}s\).

The work done by a force is measured by the product of the distance by the component of the force along the direction of the displacement.

This formula for the work is valid for forces of any origin and for motions along any trajectory.

Note that the work may be equal to zero even when forces act on a moving body.

For example, the work done by a Coriolis force is equal to zero, because such a force is perpendicular to the direction of the motion. It has no tangential component, so the work is equal to zero.

Any twist in the trajectory which is not accompanied by a change in speed requires no work, for the kinetic energy does not change under such conditions.

Can work be negative? Of course, if the force is directed at an obtuse angle to the motion, then it does not help but hinders the motion. The tangential component of the force in the direction of the motion will be negative. In this case we do say that the force performs negative work. The force of friction always slows down a motion, i.e. does negative work.

On the basis of the increase in kinetic energy, one can only judge the work done by the resultant force.

As for the work done by the individual forces, we should compute them as the products \(f_{t}s\). An automobile is moving uniformly along a highway. There is no increase in kinetic energy, so the work done by the resultant force is equal to zero. But the work done by the motor is, of course, not equal to zero—it is equal to the product of the thrust of the motor by the distance covered, and is fully compensated by the negative work done by the force of friction and resistance.

Using the concept of “work”, we can describe more briefly and clearly the interesting peculiarities of the gravitational force with which we have just become acquainted. If a body goes from one place to another under the action of gravity, its kinetic energy will change. This change in kinetic energy is equal to the work \(A\). But we know from the law of conservation of energy that an increase in kinetic energy takes place at the expense of a decrease in potential one.

Therefore, the work done by gravity is equal to the decrease in potential energy: \[W = U_{1} - U_{2}\] It is obvious that a loss (or gain) of potential energy, and hence an increase (or decrease) in kinetic energy, will be the same, regardless of the path along which a body moved. This implies that the work performed by gravity does not depend on the form of the path. If a body went from the first point to the second with an increase in kinetic energy, it will go from the second point to the first with a decrease in kinetic energy by exactly the same amount. Moreover, it makes no difference whether or not the form of the path “there” coincides with the form of the path “back”. Hence, the work “there” and “back” will also be identical. And if the body takes a long trip with the initial and end points of its path coinciding, the work will be equal to zero.

Imagine a canal whose form is as fantastic as possible, through which a body slides without friction. Let us send it off on a trip from the highest point. The body rushes downwards gathering speed. At the expense of the kinetic energy so obtained, the body will surmount ascents and return finally to the station where it departed. With what speed? With the same, of course, with which it left the station. Its potential energy will return to its previous value. But if so, then its kinetic energy could neither have decreased nor increased. Hence, the work is equal to zero.

Not for all forces is the work done along a circular (physicists say: a closed) path equal to zero. There is no need to prove that the longer the path, the greater will be the work performed by friction, for example.