When considering the movements of bodies and their systems in space, one often has to calculate the work of various forces. In this article we will give a definition of mechanical work in physics, explain how it is related to energy, and also give examples of solving problems on this topic.
What is the difference between energy and work?
When studying work in physics (grade 9 of secondary schools), many students confuse this value with energy. You can understand this: after all, both characteristics are determined in joules. However, energy is a fundamental characteristic. It cannot appear or disappear, but is only capable of passing into different states and forms. This is the essence of the law of its conservation in an isolated system. Work, on the other hand, is one of the forms of energy realization that leads to the spatial movement of bodies.
So, when a gas is heated, its internal energy increases, that is, the system gets the opportunity to do some mechanical work due to it. The latter will occur when the gas begins to expand, increase its volume.
Strict definition of work in physics
A strict definition in physics is one that requires a clear mathematical justification. As applied to the quantity under consideration, one can say the following: if a certain force F¯ acts on the body, as a result of which it begins to move to the vector S¯, then the quantity A is called work
A = (F¯ * S¯)
Since A is a scalar quantity, the parentheses on the right side of the equality indicate that both vectors are scalar multiplied.
An important fact follows from the written expression: if the force acts perpendicular to the displacement, then it does not perform work. So, many schoolchildren, when deciding on the physics of tests in the 10th grade, for example, make a frequent mistake. They believe that moving horizontally heavy loads is difficult precisely because of gravity. As the formula of work shows, the force of gravity during horizontal movement performs zero work, since it is directed vertically downward. In fact, the difficulty of moving a heavy load is associated with the action of the friction force, which is directly proportional to the force of gravity.
The expression for A in explicit form can be written as follows:
A = F * cos (φ) * S
The product F * cos (φ) is the projection of the force vector onto the displacement vector.
Work and efficiency
Everyone knows that creating a mechanism that translates all the energy expended into useful work is impossible in practice. In this regard, the concept of coefficient of performance (COP) was introduced . It is easy to calculate if you use the following expression:
Efficiency = A p / A s * 100%
Here A p , A s - useful and expended work, respectively. Moreover, A s is always greater than A p , so the efficiency is always less than 100%. For example, an internal combustion engine has an efficiency in the range of 25-40%. These figures indicate that most of the fuel during combustion is spent on heating the environment, and not on the movement of the car.
In the vast majority of cases, the inability to obtain efficiency = 100% is associated with the constant presence of friction forces. Even in such a simple mechanism as a lever, these forces acting in the area of support lead to a decrease in efficiency to 80-90%.
Further in the article, we will solve a couple of problems on the topic discussed.
The task with the body on an inclined plane
A body weighing 4 kg moves vertically up an inclined plane. The angle of its inclination relative to the horizon is 20 o . An external force acts on the body, which is 80 N (it is directed horizontally), as well as a friction force, which is 10 N. It is necessary to calculate the work of each of the forces and the total work if the body moved along a plane of 10 meters.
Before starting to solve the problem, we recall that, in addition to the indicated forces, the body is also affected by gravity and support reactions. The latter can not be considered, since its work will be zero. Gravity does negative work as the body moves upward along the sloping.
First, we calculate the work of an external force F 0. It will be:
A 0 = F 0 * S * cos (20 o ) = 751.75 J.
Note that the calculated work will be positive, since the vector of external force has an acute angle with the direction of movement.
The work of gravity F g and friction F f will be negative. We calculate them taking into account the angle of inclination of the plane and the direction of movement of the body:
A 1 = -F g * S * sin (20 o ) = -m * g * S * sin (20 o ) = -134.21 J;
A 2 = -F f * S = -10 * 10 = -100 J.
The total work of all forces will be equal to the sum of the calculated values, that is:
A = A 0 + A 1 + A 2 = 751.75 - 134.21 - 100 = 517.54 J.
This work is spent on increasing the kinetic energy of the body.
A task with a complex dependence of strength
It is known that a material point moves along a straight line, changing its coordinates from x = 2 to x = 5 m. In the process of movement, a force F exerts its influence on it, which changes according to the following law:
F = 3 * x 2 + 2 * x - 5 N.
Assuming that F acts along the line of movement of the point, it is necessary to calculate the work that it does.
Since the force is constantly changing, it’s not possible to use the formula for A. written in the article in our forehead. To calculate this value, we will proceed as follows: we calculate the work dA on each elementary segment of the path dx, and then add all the results. Arguing like this, we come to the integral formula for working in physics:
A = ∫ x (F * dx).
Now it remains to calculate this integral for our case:
A = ∫ 5 2 ((3 * x 2 + 2 * x - 5) * dx) = (x 3 + x 2 - 5 * x) | 5 2 = 123 J.
We got the result in joules, because the x coordinate is expressed in meters, and the force F is in newtons.