Set 6 Problem number 10

The graph below depicts the net force parallel to the x axis on an object vs position along the x axis, with force in Newtons and position in meters.. The gridlines depict units of .8 Newtons in the vertical direction and 5 meters in the horizontal direction.

What is the area under the graph for the displacement depicted, and what is the specific meaning of this area?

Solution

The area under the curve could be broken into tiny trapezoids with altitudes representing forces in Newtons and widths representing displacements in meters. The average of the altitudes of each trapezoid represents the approximate average force on that interval, and the width of the trapezoids represent the displacements over which this approximate force is sustained.

The area of a trapezoid therefore represents a product of average force and displacement, which represents the work done on the object, equal to the change in the object's KE.
When summed the total area of the trapezoids therefore represents the approximate work done on the object, equal to the change in its KE, over the entire displacement.
In the limit as the widths of the trapezoids approach zero the total area represents the exact work on or change in KE of the object.

We therefore estimate the area under the curve.

We find the area by a trapezoidal approximation using several well-chosen trapezoids, or alternatively by counting the rectangles under the curve (the area of each rectangle representing the displacement .8 Newtons * 5 meters = 4 Newton meters = 4 kg m/s).

University Physics note that the work would be obtained by integrating the function represented by the graph between the extreme times represented by the graph.