In this section it is study what is the effect of
applying a force over an object along a given displacement,
. Let us consider a
box of mass m being pulled over a horizontal frictionless surface
by a force as shown
in the drawing. The box is pulled along the entire displacement
shown in the drawing.
The diagram does not include the vertical forces acting on the
box; that is, the weight of the box and the normal force exerted by the
surface on the box because these two forces cancel each other in this
case. Since the applied force,
, is the only force
pointing in the horizontal direction (remember that there is not
friction present between the surfaces) this force is the net force
acting on the box.
Based on the previous situation, a physical quantity,
called work is defined:
. Here, the work is
defined as the product of the multiplication of the magnitude of the
force and the magnitude of the displacement.
This drawing shows the case when the force,
, is not parallel to
the displacement, .
In this case, the work is NOT defined as the product of the magnitudes
but it is defined in terms of the component of the force parallel to the
displacement. In fact, the force
can be broken into
its parallel,
, and
perpendicular components,
.
The last drawing of the sequence shows the angle that is used for
defining these components,
.
Therefore, the definition of work done by a constant force is
where
is the
component of the force parallel to the displacement and
is the magnitude of the
displacement.
The parallel component of the force can be calculated using
trigonometry, see the definition of the angle
.
This component is
. The last
expression for the work done by the applied force corresponds to the scalar
product between two vectors,
Notices the following characteristics of this definition:
The component of the force perpendicular to the
displacement,
, does not
contribute to the work done by the force.
The work is NOT a vector quantity. It does not have
direction. Nevertheless, the work can be positive or negative depending on
the sign of
. Most
scalar quantities that have been studied so far are positives. Therefore,
especial attention must be paid when calculating work.
This definition is only valid when the magnitude and
direction of the applied force are constant.
The unit of work is derived from the units of force and
displacement, Unit of Work = (Newton) (Meter) = Nm. This
unit is called Joule (J); therefore,
The work done by a constant force can also be calculated
using a graphical representation as shown below.
The force acting on an object along the
displacement
is represented in the graph
of the left. In this case,
the force acting on the object is assumed to be in the same
direction as the displacement.
Moreover, the magnitude and direction of
the force are constant. Graphically, such a force can be represented as
the first graph shown in the left,
.
The area under the line of this graph is represented in the second
graphically
illustration:
.
Such an area is simple given by
which, on the other side, is exactly the value of the work done by the force,
.
Therefore, the area under the curve (line in this case) of the force
versus the displacement correspond to the work done by the force along
this displacement,
.
This result can be used to resolve situations where the force is not
constant such as
pulling on
a spring.
When different forces are applied to an object, different works can be
calculated. For each force acting on an object, the corresponding work can be
calculated,
Enter the correct
amount of work including the sign in the corresponding
textbox. Check your response
by clicking anywhere in this page. Show question again by
clicking on the corresponding most left drawing.
J
The box is moving to the
right along a displacement of 10m over a frictionless surface.
At the same time, a force of 5N is
pulling on the box in the same direction.
Since the force and the displacement are
in the same direction, the correct answer is obtained by
just multiplying the force by the displacement,
.
J
The box is moving toward the
right along a displacement of 10m over a frictionless surface.
Also, a force of 5N is
pulling on the box in the opposite direction as the
displacement.
The work can
be calculated following the sign convention for vectors; that is,
considering vectors pointing to the right to be positives and
vectors pointing to the left to be negatives. Therefore, the
work in this case is
.
J
The box is moving toward the
left along a displacement of 10m over a frictionless surface.
Also, a force of 5N is
pulling on the box in the opposite direction as the
displacement.
The work can
be calculated following the sign convention for vector; that is,
considering vectors pointing to the right to be positives and
vectors pointing to the left to be negatives. Therefore, the
work in this case is
.
J
The box is moving toward the
left along a displacement of 10m over a frictionless surface.
Also, a force of 5N is
pushing on the box in the same direction as the
displacement.
Again, the work can
be calculated following the sign convention for vector; that is,
considering vectors pointing to the right to be positives and
vectors pointing to the left to be negatives. Therefore, the
work in this case is
.
J
The box is moving toward the
right along a displacement of 10m over a frictionless surface.
Also, a vertical force of 5N is
applied to the box in the direction shown in the diagram.
In this case, the work can
be calculated accordingly with the definition stated above,
,
but with
,
,
and
.
Since the ,
the work done by this force is
.
J
The box is moving toward the
right along a displacement of 10m over a frictionless surface.
Also, a force of 5N is
applied to the box in the direction shown in the diagram.
Find the resulting work done by the force rounded to the nearest
unit.
In this case, the work can
be calculated accordingly with the definition stated above,
,
but with
,
,
and
.
Since the
,
the work done by this force is
.
J
The box is moving toward the
right along a displacement of 10m over a frictionless surface.
Also, a force of 5N is
applied to the box in the direction shown in the diagram.
Find the resulting work done by the force rounded to the nearest
unit.
In this case, the work can
be calculated accordingly with the definition stated above,
,
but with
,
,
and
.
Since the
,
the work done by this force is
.
Another interesting situation occurs when it is considered
the work done by a person walking along a horizontal floor carrying a given load
such as a backpack or briefcase. In this case, if the walker maintains the same pace, the
work done by the walker is zero no matter how heavy is the load. Nevertheless,
if the walker accelerates in any direction, during the acceleration the work
done by the walker is non-zero. As shown in the drawn, while carrying an object
with constant velocity, the force exerted by the women,
, is
perpendicular to her displacement implying that the work done by her is
.
However, this is not the case when the person climbs a
set of stairs.
In this case, when there is not acceleration, the
force exerted by the person is vertical upward and of the same magnitude
as the weight of the briefcase. Nevertheless, this force has a component
along the direction of the displacement (even when there is not
acceleration); notice that, in the drawing, the displacement is
represented by . Thus
the work done by the person when carrying the briefcase is
,
with .
On the other side,
which implies that
the work done by the person is
.
In addition, the force exerted by the woman is exactly the
same as the weight of the briefcase,
, which
implies that the work done by her is
.
In most cases, for a given situation, there are different
forces acting on the object of interest. After the forces have been identified,
the work done by each of the forces can be calculated. In addition, the
work
done by the net force can also be calculated.
Show
Forces
Let us consider the work done by the different forces
acting on the object when lifting it. For this simple situation,
there are two forces acting on the object plus the net force resulting
of their addition. The forces are the weight of the object,
, and
the external force applied to the object by the agent lifting the
weight. This external force has the following characteristics,
When the object is supported by the floor (normal
force), the external force just need to be sufficient to put the
object in motion.
At the beginning of the motion, the external
force must be slightly bigger than the weight of the
object because this force
must accelerate the weight in order to start lifting it. The
velocity to be reached by the weight should be very small in order
to minimize the difference between the external force and the weight
of the object.
After the object is in motion, the magnitude of
the external force match the magnitude of the weight exactly because
the object needs only to be maintain in motion with a very small
constant velocity.
When the object start reaching the upper most
point in the path in its motion toward reaching the table, the
magnitude of the applied force must be slightly smaller than the
weight of the object in order to change the direction of the
velocity of the object. This condition on the magnitude of the force
is maintain until the object reaches the wanted descending velocity.
Again, this velocity is very small.
When descending over the table, yet again the
magnitude of the force is exactly the same as the magnitude of the
weight of the object. In this part of the motion, the object is
maintained descending with very small constant velocity.
When reaching the top of the surface, the
magnitude of the external applied force is for the last time
slightly bigger than the weight of the object. Now, the external
force must slow down the object to zero velocity.
Therefore, taking in account the previous
points, the slower the speed of the object, the closer the "average"
magnitude of the external force is to the magnitude of the weight of the
object. Such a process of lifting an object is called a quasi static
process.
After the insights about the external force presented
above, the work done by this force can be calculated. In the drawing on
the left, the path of the object has been broken down into two basic
pieces, the piece when the object is moving upward and when the object
is moving downward,
.
Horizontal displacements of the block
do not contribute to the work done
by the force because the angle between those displacement and the
direction of the force is 90^{0},
. When the object is moving
upward the direction of the force is in the same direction as the
displacement,
. Contrary, when the object is moving downward, the
direction of the applied force and the direction of the displacement are
in opposite direction,
.
In general, the total displacement of an object is
the vector addition of the partial displacements of the object. In this
case,
which is represented in the drawing,
.
Therefore, the work done by the external force is
Similarly, the work done by the gravitational force
(weight) can be calculated following the diagram shown,
.
In this case, the angle between the displacement and the force
(weight) is 180^{0} and the work is
.
Notice that these two works have the same magnitude but opposite signs;
however, the force associated to the weight is always constant during
the entire processes. Comparing the weight to the external force, the
weight is a better candidate for the definition of mechanical potential
energy (add hyperlink) because the external
force must be understood with all the variations remarked above.
The net work is the work done on an object by all the forces
acting on the object or, equivalently, the work done by the net force. Thus, if
several forces,
, act on an
object, the net work done by all these forces is the work done by the net force,
, where
is the displacement of the
object.
The previous statements are valid when the forces are
constant and the displacement is in a straight line. In other cases, calculus
techniques are used to obtain the corresponding work.
The net work can also be obtained by solving the algebraic
addition of the works done by the individual forces,
Thus, the net work done in lifting the object presented above
can be obtained by just applying the previous result to the partial results
obtained at the end of the previous section,
The fact that the net work is zero, indicates that there was
not a net change in the kinetic energy of the object during the process of lifting the object.