Mechanical work - definition, basic formulas and calculation examples

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Definition Elementary work ($dA$) is a physical quantity that is equal to the scalar product of force ($\overline{F}$) and infinitesimal displacement ($d\overline{s}$) that occurs under the influence of this force:
\[ dA=\overline{F}\cdot d\overline{s}=F{\cos \alpha \ ds\ \left(1\right).\ }\]

If a body moves in a straight line along path $s$ under the influence of a constant force $\overline{F}$, then the work can be calculated as:

\[A=Fs{\cos \alpha \ \left(2\right),\ }\]

where $\alpha $ is the angle between the direction of the force and the direction of movement of the body.

What is work in physics - definition and formula

Mechanical work is the amount of energy that must be expended in order for the body to begin a uniformly decelerating motion and cover a certain distance.

In physics, mechanical work is the product of a force that acts on a body and the distance it travels under its influence:

A=F*S

In more complex cases, a third quantity appears in the formula - the cosine of the angle at which the vectors of motion and applied force are located to each other. You can find its value using the formula:

A = F * S * cosA

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How is work measured?

The physical units in which mechanical work is expressed are Joules.

There are different ways to practically measure it, depending on the type of movement produced. In this case, the value of the force in Newtons and the distance in meters are substituted into the work formula. The angle between vectors is measured in mathematical units - degrees.

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Work of friction force

Under the conditions existing on Earth, any moving body is affected by the force of friction, which slows down its movement. Most often this is the friction of the surface on which the object is moving. This is obvious from the fact that when a constant force is applied to a body, its speed will be variable.

Therefore, there must be another force opposing it - and this is the force of friction. If the coordinate system is chosen in the direction of body movement, then its numerical value will be negative.

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Positive and negative work

The numerical value of the work performed by a force can become negative if its vector is opposite to the velocity vector.

In other words, force can not only give the body speed to make a movement, but also impede the movement already being made. In this case, it will be called counteracting.

Useful or wasted work

A body performing the same action has two meanings of work. The first of them, useful, is calculated using the usual formula.

The second, spent, in its concept does not have a general formula for calculation and is measured practically. This difference between the work done in reality and the work that should have been done in theory is equal to the coefficient of performance - efficiency. It is calculated like this:

Efficiency = A useful / A expended

,

and is expressed as a percentage. Efficiency is always less than 100.

Power

The average amount of work done per unit of time (second) characterizes such a quantity as power. The formula for calculating it looks like this:

P = A/t

As a work, you can substitute a well-known formula to calculate it, depending on the situation. The answer will be expressed in Watts.

However, with uniform motion, you can use another formula:

P = F * v

By substituting instantaneous speed instead of normal speed, you can get the value of instantaneous power.

General information.

The concept of work is part of the formulation of the first and second laws of thermodynamics.

The work is usually designated by the letter - A. The units of work naturally correspond to the units of energy.

The concept of work is rarely used when solving applied problems. More often they use the concept of power (power of engines, power of technical devices, etc.) defined on the basis of work done per unit of time (energy used per unit of time).

Examples of problem solving

Let's consider several simple problems of finding mechanical work.

Problem 1

What work is done by a lifting mechanism that lifts a ten-kilogram block to a height of 50 meters?

Solution:

In order to lift a body, it is necessary to overcome the force of gravity acting on it. That is, the F with which the block is lifted is equal to the one with which it is attracted to the ground. Since the latter is equal to m * g, then to find the final result you only need one modified version of the standard formula mentioned above: A = S * m * g.

Using simple mathematics, we find the numerical answer:

A = 50 m * 10 kg * 10 N/kg;

A = 5000 J.

Answer: 5000 J.

However, it is not always about gravity.

Problem 2

What work is done by the elastic force when a spring with a stiffness of 10 N/m, compressed by 20 cm, returns to its original state? The system is closed, there are no external forces acting on the spring.

Solution:

First you need to find the elasticity F itself, which does the work. Its formula is F = x * |k|, where x is the length by which the spring is compressed or stretched, and k is its stiffness coefficient. The displacement of the spring is equal to its deformation, and therefore the final formula in this case will look like this: A = S * x * k = x * x * k = x^2 * k.

Next, using elementary calculations, we calculate the answer:

A = (0.2 m)^2 * 10 N/m = 0.04 * 10 = 0.4 J.

Answer: 0.4 J.

But in all problems on this topic, the trajectory of the body’s movement is straight.

Problem 3

Calculate the force acting on the wheel if it takes 10 kJ to complete a full rotation. The diameter of the disk is 40 cm, and the thickness of the tire is 10 cm.

Solution:

In this case, we need to find not A, but F, but this can be done using the same formula. Let's take a point on the surface of the wheel. Let us assume that during rotational motion its vector will be opposite to the vector of force application, which means that the cosine in the formula can again be neglected. Thus, in one revolution of the wheel the point will travel a distance equal to the circumference, which can be calculated as 2πr or πd. The diameter of the circle can be found from the data provided: it is equal to the sum of the rim diameter and twice the tire thickness, that is, 40 cm + 2 * 10 cm = 40 cm + 20 cm = 60 cm = 0.6 m.

Now that we can calculate the distance, we have all the data we need to start finding the force.

The work formula for this case will be: A = F * π * d, then the force, accordingly, can be expressed as F = A / (π * d).

In this case:

F = 10 kJ / (3.14 * 0.6 m) = 10000 J / 1.884 m = ~5308 N.

Answer: 5308 N.

In conclusion, we will solve the most complex version of the problem, which includes everything mentioned above.

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