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 dot 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 lifts the weight, he works against gravity, but when he keeps the weight still, from a physics point of view, he is not working. source: needpix.com
When the force is not constant, we must analyze the work done when displacements are very small or differential. In this case, if it is considered a starting point for point A and as an arrival for B, the total work will be obtained by adding all contributions to it. This is equivalent to calculating the following integral:
Variation in system energy = Work performed by external forces
ΔE = W ext
When energy is added to the system, W> 0 and when energy is subtracted W <0. Now, if ΔE = 0, it could mean that:
-The system is isolated and there are no external forces acting on it.
-There are external forces, but they are not working in the system.
As the change in energy is equivalent to the work done by external forces, the SI energy unit 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 a dot product between two vectors:
W = F ● l = θ FLCOS
Where F is the magnitude of the force, l is the magnitude of the displacement, and θ is the angle that exists 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.
Rewriting the work as follows:
W = (F. cos θ). me
We can say that only the component of force parallel to the displacement: F. cos θ e is capable of performing the work. If θ = 90º then cos θ = 0 and the work would be null.
Therefore, it is concluded that forces perpendicular to the displacement do not perform mechanical work.
In the case of figure 2, neither the normal force N nor the weight P work, since both are perpendicular to 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, as it has the same direction of movement.
If cos θ = 1 , force and displacement are parallel and work is maximum.
In the case of cos θ <1, the force is not in favor of the movement and the work is negative.
When cos θ = -1 , the force is completely opposite to displacement, like kinetic friction, whose effect is to restrict the object on which it acts. So the work is minimal.
This is consistent 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.
Net work W net is defined as the sum of the work done by all forces acting in the system:
net W = ∑W i
We can then conclude that, to ensure the existence of liquid mechanical work, it is necessary that:
-Activate external forces on the object.
-These forces are not all perpendicular to the displacement (cos θ ≠ 0).
-The work carried out by each force is not cancelled.
-There’s a shift.
Examples of mechanical work
Whenever it is necessary to put an object in motion, it is necessary to carry out mechanical work. For example, pushing a refrigerator or heavy chest across a horizontal surface.
Another example of a situation where mechanical work is required is changing the speed of a moving ball.
-You need to work to raise an object to a certain height above the ground.
However, there are equally common situations in which no work is performed, although appearances indicate otherwise. We said that to raise an object to a certain height, we have to work, so we carry the object, lift it above our head, and keep it there. We’re working?
Apparently, yes, because if the object is heavy, the arms will get tired soon after, no matter how hard you do it, no work is being done from a physics point of view. Why not? 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 who spins 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. So he does not perform mechanical work, despite favoring movement.
The Kinetic Energy Theorem of Work
The kinetic energy of the system is what it has by virtue of its motion. If m is the mass and v is the speed of motion, the kinetic energy is denoted by K and is given by:
K = ½ mv 2
By definition, the kinetic energy of an object cannot be negative, as the mass and square of velocity are always positive quantities. 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 assume that the mass remains constant, although this is not always the case. This requires net work on the system, so:
net W = Δ K
This is the kinetic energy theorem of work. Says:
The net work is equivalent to the change in the kinetic energy of the system
Note that although K is always positive, ΔK can be positive or negative because:
ΔK = final K – initial K
If K end > K initial, the system gained power and ΔK> 0. Conversely, if K end < K initial , the system transferred power.
Work performed to stretch a spring
When a spring is stretched (or compressed), the work needs to be done. That work is stored in the spring, allowing it to work in, say, a block 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 this displacement. The proportionality constant depends on how the spring is: soft and easily deformable or rigid.
This strength is given by:
F r = -kx
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 displacement.
Figure 3. A compressed or stretched spring works on an object attached to its end. Source: Wikimedia Commons.
If the spring is compressed (left in figure), the block at the end will move to the right. And when the spring is stretched (to the right), the block wants to move to the left.
To compress or stretch the spring, some external agent must do the work and, as it is a variable force, to calculate this work, the definition given at the beginning must be used:
It is very important to note that this is the work performed by the external agent (a person’s hand, for example) to compress or stretch the spring. That’s why the minus sign doesn’t appear. And since the positions are square, it doesn’t matter if they’re compressions or stretches.
The work that spring will do on the block is:
W spring = -W ext
Exercises
Exercise 1
The block in Figure 4 has mass M = 2 kg and slides along the sloped plane without friction, with α = 36.9 º. Assuming that it is allowed to slide from the rest of the plane, whose height is h = 3 m, find the speed at which the block reaches the base of the plane, using the kinetic energy of work theorem.
Solution
The free-body diagram shows that the only force capable of working on the block is the weight. More accurate: the weight component along the x-axis.
The distance traveled by the block on the plane is calculated by trigonometry:
d = 3 / (cos 36.9º) m = 3.75 m
Weight W = (Mg). d. cos (90-α) = 2 x 9.8 x 3.75 x cos 53.1 º J = 44.1 J
For the kinetic work energy theorem:
net W = Δ K
net W = weight W
ΔK = ½ Mv f 2 – ½ Mv or 2
As it is released from rest, v o = 0 , therefore:
net W = ½ Mv f 2
Exercise 2
A horizontal spring, whose constant is k = 750 N/m, is attached at one end to a wall. One person presses the other end to 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 force applied by the person is:
F = kx = 750 N / m. 5 x 10 -2 m = 37.5 N.
b) If the end of the spring was originally at x 1 = 0, to take it from there to the final position x 2 = 5 cm, it is necessary to carry out the following work, according to the result obtained in the previous section:
W ext = ½ k (x 2 2 – x 1 2 ) = 0.5 x 750 x 0.05 ( 2 -0 2 ) J = 0.9375 J.
