MECHANICAL WORK: WHAT IS IT, CONDITIONS, EXAMPLES, EXERCISES - PHYSICAL - 2025

The mechanical work is defined as the change in the energy state of a system, caused by external forces such as gravity or friction. The units of mechanical work in the International System (SI) are newton x meter or joules, abbreviated by J.

Mathematically it is defined as the scalar product of the force vector and the displacement vector. If F is the constant force and l is the displacement, both vectors, the work W is expressed as: W = F l

Figure 1. While the athlete is lifting the weight, he does work against gravity, but when he keeps the weight immobile, from the point of view of Physics he is not doing work. source: needpix.com

When the force is not constant, then we must analyze the work done when the displacements are very small or differential. In this case, if point A is considered as the starting point and B as the arrival point, the total work is obtained by adding all the contributions to it. This is equivalent to calculating the following integral:

Variation in the energy of the system = Work done by external forces

When energy is added to the system, W> 0 and when energy is subtracted W <0. Now, if ΔE = 0, it can mean that:

-The system is isolated and there are no external forces acting on it.

-There are external forces, but they are not doing work on the system.

Since the change in energy equals the work done by external forces, the SI unit of energy is also the joule. This includes any type of energy: kinetic, potential, thermal, chemical, and more.

Conditions for mechanical work

We have already seen that work is defined as a dot product. Let's take the definition of work done by a constant force and apply the concept of the dot product between two vectors:

Where F is the magnitude of the force, l is the magnitude of the displacement and θ is the angle between the force and the displacement. In figure 2 there is an example of an inclined external force acting on a block (the system), which produces a horizontal displacement.

Figure 2. Free-body diagram of a block moving on a flat surface. Source: F. Zapata.

Rewriting the work in the following way:

We can say that only the component of the force parallel to the displacement: F. cos θ is capable of doing work. If θ = 90º then cos θ = 0 and the work would be zero.

Therefore it is concluded that the forces perpendicular to the displacement do not do mechanical work.

In the case of Figure 2, neither the normal force N nor the weight P do work, since they are both perpendicular to the displacement l.

The signs of work

As explained above, W can be positive or negative. When cos θ> 0, the work done by the force is positive, since it has the same direction of motion.

If cos θ = 1, the force and displacement are parallel and the work is maximum.

In case cos θ <1, the force is not in favor of the motion and the work is negative.

When cos θ = -1, the force is completely opposite to the displacement, such as kinetic friction, whose effect is to slow down the object on which it acts. So the work is minimal.

This agrees with what was said at the beginning: if the work is positive, energy is being added to the system, and if it is negative, it is being subtracted.

The net work W _net is defined as the sum of the works done by all the forces acting on the system:

Then we can conclude that to guarantee the existence of net mechanical work it is necessary that:

-External forces act on the object.

-Said forces are not all perpendicular to the displacement (cos θ ≠ 0).

-The jobs done by each force do not cancel each other.

-There is a displacement.

Examples of mechanical work

-Whenever it is required to put an object in motion starting from rest, it is necessary to do mechanical work. For example pushing a refrigerator or a heavy trunk on a horizontal surface.

-Another example of a situation in which it is necessary to do mechanical work is to change the speed of a moving ball.

-It is necessary to do work to raise an object to a certain height above the floor.

However, there are equally common situations in which work is not done, although appearances indicate otherwise. We have said that to lift an object to a certain height you have to do work, so we carry the object, raise it above our head, and hold it there. Are we doing work?

Apparently yes, because if the object is heavy the arms will get tired in a short time, however, no matter how hard it is, no work is being done from the point of view of Physics. Why not? Well, because the object is not moving.

Another case in which, despite having an external force, it does not perform mechanical work is when the particle has a uniform circular motion.

For example a child spinning a stone tied to a string. The string tension is the centripetal force that allows the stone to rotate. But at all times this force is perpendicular to the displacement. Then he does not perform mechanical work, although it favors movement.

The work-kinetic energy theorem

The kinetic energy of the system is that which it possesses by virtue of its movement. If m is the mass and v is the speed of motion, the kinetic energy is denoted by K and is given by:

By definition, the kinetic energy of an object cannot be negative, since both the mass and the square of the velocity are always positive quantities. The kinetic energy can be 0, when the object is at rest.

To change the kinetic energy of a system, its speed must be varied - we will consider that the mass remains constant, although this is not always the case. This requires doing net work on the system, therefore:

This is the work - kinetic energy theorem. It states that:

Note that although K is always positive, ΔK can be positive or negative, since:

If _final K > _initial K the system has gained energy and ΔK> 0. On the contrary, if _final K < _initial K, the system has given up energy.

Work done to stretch a spring

When a spring is stretched (or compressed), work must be done. This work is stored in the spring, allowing the spring to do work on, say, a block that is attached to one of its ends.

Hooke's law states that the force exerted by a spring is a restitution force - it is contrary to displacement - and also proportional to said displacement. The constant of proportionality depends on how the spring is: soft and easily deformable or rigid.

This force is given by:

In the expression, F _r is the force, k is the spring constant, and x is the displacement. The negative sign indicates that the force exerted by the spring opposes the displacement.

Figure 3. A compressed or stretched spring does work on an object tied to its end. Source: Wikimedia Commons.

If the spring is compressed (to the left in the figure), the block at its end will move to the right. And when the spring is stretched (to the right) the block will want to move to the left.

To compress or stretch the spring, some external agent must do the work, and since it is a variable force, to calculate said work, we must use the definition that was given at the beginning:

It is very important to note that this is the work done by the external agent (a person's hand, for example) to compress or stretch the spring. That is why the negative sign does not appear. And since the positions are squared, it doesn't matter if they're compressions or stretches.

The work that the spring will in turn do on the block is:

Exercises

Exercise 1

The block in figure 4 has mass M = 2 kg and slides down the inclined plane without friction, with α = 36.9º. Assuming that it is allowed to slide from rest from the top of the plane, whose height is h = 3 m, find the speed with which the block reaches the base of the plane, using the work-kinetic energy theorem.

Figure 4. A block slides downhill on an inclined plane without friction. Source: F. Zapata.

Solution

The free-body diagram shows that the only force capable of doing work on the block is weight. More accurate: the component of weight along the x-axis.

The distance traveled by the block on the plane is calculated using trigonometry:

By work-kinetic energy theorem:

Since it is released from rest, v _o = 0, therefore:

Exercise 2

A horizontal spring, whose constant is k = 750 N / m, is fixed at one end to a wall. A person compresses the other end a distance of 5 cm. Calculate: a) The force exerted by the person, b) The work he did to compress the spring.

Solution

a) The magnitude of the force applied by the person is:

b) If the end of the spring is originally at x ₁ = 0, to take it from there to the final position x ₂ = 5 cm, it is necessary to do the following work, according to the result obtained in the previous section:

References

1. Figueroa, D. (2005). Series: Physics for Science and Engineering. Volume 2. Dynamics. Edited by Douglas Figueroa (USB).
2. Iparraguirre, L. 2009. Basic Mechanics. Natural Sciences and Mathematics Collection. Free online distribution.
3. Knight, R. 2017. Physics for Scientists and Engineering: a Strategy Approach. Pearson.
4. Physics Libretexts. Work-energy theorem. Recovered from: phys.libretexts.org
5. Work and Energy. Recovered from: physics.bu.edu
6. Work, energy and power. Retrieved from: ncert.nic.in