Work Done by a Constant Force

 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:

1. The component of the force perpendicular to the displacement, , does not contribute to the work done by the force.

2. 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.

3. This definition is only valid when the magnitude and direction of the applied force are constant.

 Unit of Work

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.

 Understanding Work

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 900, . 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 1800 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.

 by Luis F. Sáez, Ph. D. Comments and Suggestions: LSaez@dallaswinwin.com