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Liceo Scientifico Isaac Newton – Roma
work
in accordo con il
Ministero dell’Istruzione, Università, Ricerca
e sulla base delle
Politiche Linguistiche della Commissione Europea
percorso formativo a carattere
tematico-linguistico-didattico-metodologico
scuola secondaria di secondo grado
professor
Serenella Iacino
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Work
In daily life we use the term work to indicate an activity or an effort, but in Physics
this term has a different meaning .
First of all in Physics Work is always done by forces, in fact a force acting upon an
object does work when the initial point of application of the force is displaced.
For example, we do work when we push a trolley in an airport.
But if we stand still lifting a bag, we are not working even if we feel tired!
Generally, if F is a force acting upon an object, the mechanical work done by the
force is the product of the component of the force that points in the direction of the
displacement, and the modulus of the same displacement where Fcos θ comes
from the first theorem of trigonometry for right-angle triangles.
So, if F is the magnitude of the force acting upon an object, θ is the angle
between F and the displacement s, and if Fcos θ is the component of the force in
the direction of the displacement s, the work is the dot product between the
force and the displacement.
This formula is true only if the force and angle remain constant.
Its unit of measure is joule in honour of the English Physicist James Prescott Joule
who studied energy and its transformation.
Work depends on the angle between the force F and the displacement vector s, so
we have three possibilities:
1. if the angle THETA is less than 90°, the mechanical work is positive;
2.
if the angle THETA is greater than 90°, the mechanical work is negative;
3. if the force and the displacement are perpendicular to each other, then the
work is zero;
When many forces act upon an object, we can obtain the total work by two
equivalent methods:
We can determine the work done by each force acting upon an object and then we
add them to obtain the sum of work;
Let’s suppose that a man pulls a case on a flat surface using a rope inclined at
angle TETA to the floor, acting as a constant force F upon the case.
The floor develops a friction force Fa; we want to know what is the sum of work
done on the case, knowing that the force of gravity P and the normal force N also
act upon it.
So the sum of work is as shown.
We can determine the resultant of all forces acting upon an object and then we
determine the work of the resultant.
Let’s draw the free-body diagram and split the forces into the x and y components
on a convenient Cartesian plane that has the x axis parallel to the flat surface.
From Newton’s second law we have…..
So the magnitude of the resultant of all forces is……….
Thus, the angle between the resultant and the displacement is equal to zero, and
the total work is as shown .
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We can represent graphically on a Cartesian plane the work of a constant force F
that is parallel to the displacement s and has the same direction, placing the
displacement along the x axis and the force along the y axis.
In this case THE WORK IS ………and is represented by the area of the rectangle
having base s1 and heigth F1.
But if the force F, represented by the function f(x) , changes over time, we can
divide the displacement s1 into small steps of length Δs , in each of which the
force can be considered constant; then we can draw rectangles having base Δs1
approximating the curve of the force: the greater the number of rectangles, the
better the approximation is.
When the number of the rectangles is infinite, then the work is represented by the
area under the curve between 0 and s1 .
So the general definition of mechanical work is given by the following integral.
For example, the force of gravity P is a constant force and acts vertically
downwards.
If an object, having mass m, falls from a height h, the angle between the gravity
force and the displacement is zero, so the work done by this force is the following.
Let’s make another example:
During a football match, a ball of mass m is kicked up, reaching a height h and
then falls to the ground. Calculate the work done by the gravity force P during the
ascent and during the descent.
we can see the ascent work is the following and the descent work is as shown .
The elastic force is the force exerted by a stretched or compressed spring, and is a
variable force.
If we apply a force F on a spring of length L at rest, we stretch the spring by
extension x.
Now we split the force F into its components on a convenient Cartesian plane that
has the x axis parallel to the spring.
The component F is directly proportional to the extension x of the spring from its
equilibrium position, according to the linear relationship F = Kx, where K is the
spring constant. The higher the K value, the stiffer the spring is.
From Newton’s third law, the spring reacts to the action of the force F, exerting
another force Fel, having the same magnitude but opposite direction.
This force is called the elastic force and pulls the spring back towards the wall.
The relationship is Hooke’s law and is valid only for those objects which, once the
stress is removed, return to their original length, without permanent deformation.
In the same way, if the spring is compressed so that F is directed towards the left ,
the spring reacts exerting an elastic force directed towards the right.
The work done by a spring can be calculated geometrically, after drawing the graph
of the elastic force on a Cartesian plane.
These graphs are represented by a straight line passing through the origin of the
Cartesian plane and having the gradient +K, if the spring is compressed, or “–K”, if
the spring is stretched.
Grafically the work is given by the area of the triangle.
From a mathematical point of view we have the following formula.
In Physics, Energy is the ability to perform work.
Suppose we apply a force F to a stationary object of mass m and that this force
changes its velocity;
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this object will move in a uniformly accelerated motion in the direction of force.
If in time t, its displacement is s and its velocity is v, remembering the laws of
uniformly accelerated motion. The work done by the force F , to accelerate the
object from rest to velocity v is Kinetic Energy, which can be called just K and is
the energy of motion.
It’s measured in joules and is a scalar quantity.
If at the beginning, the object moves with velocity vi then, the force F gives the
object an acceleration “a” that changes its velocity to vf . So, if in time t its
displacement is s, from the laws of the uniformly accelerated motion we have…….
Where substituting t in the second equation, we have……..
Thus the work done by the force F is……….
This relationship is called the work – energy theorem.
We observe that if the work is positive, then the kinetic energy increases when the
object goes from the initial to the final position;
If the work is negative, then the kinetic energy decreases.
If the work is zero, the kinetic energy doesn’ t change, so the velocity is constant.
Let’s make an example, a man pushes a supermarket trolley of mass m = 10 kg,
for two metres, applying a force F the magnitude of which is 100 N; the friction
force between the floor and the trolley is 30 N.
Knowing that the force of gravity P and the normal force N also act upon the
object, we draw the free-body diagram and we determine the work done by each
force.
The work done by force F and by friction Fa is as shown.
Applying the work – energy theorem we have the following relationship from which
we have the final velocity of the trolley.
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