Solved Problem on Work and Energy

A body moves in a straight trajectory, the graph of force versus displacement of the body is shown in the following figure.

a) Between what points is there no force acting on the body, between which points does the force produce movement, and between which points does the force resist movement?
b) What is the work done by the force between points 0 and 60 m?

Solution:

a) Between points 25 and 40 m, the force is equal to zero, F = 0, and there is no force acting on the body.
Between the points 0 and 25 m, the force is positive, F>0, and it is in the same direction as the displacement (e.g., the force of the motor of a car).
Between the points 40 and 60 m, the force is negative, F<0, it is in the opposite direction of the displacement, and the force is resistive (e.g, the force exerted by the brake of a car).

Note: e.g., is the abbreviation of the expression in Latin "exempli gratia", which means "for example."

b) The work done by the force F between 0 and 25 m will be numerically equal to the trapezoid area under the curve in the graph and the x-axis, highlighted in gray (Figure 1). The trapezoid area is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {A=\frac{(B+b)h}{2}} \end{gather} \]

Figure 1

using the values from the graph

\[ \begin{gather} _{\small F}W_{0}^{25}\;\overset{\mathrm{N}}{=}A=\frac{[25+(10-5)]\times200}{2}\\[5pt] _{\small F}W_{0}^{25}=3000\;\mathrm J \end{gather} \]

Between points 25 and 40 m, and the force on the body is equal to zero, F = 0, the work will be zero.

\[ \begin{gather} _{\small F}W_{25}^{40}=0 \end{gather} \]

Between points 40 and 60 m, the work will be numerically equal to the area of the triangle under the x-axis and the curve, highlighted in gray (Figure 2). The triangle area is given

\[ \begin{gather} \bbox[#99CCFF,10px] {A=\frac{B\;h}{2}} \end{gather} \]

Figure 2

using the values from the graph

\[ \begin{gather} _{\small F}W_{40}^{60}\;\overset{\mathrm{N}}{=}A=\frac{(60-40)\times(-200)}{2}\\[5pt] _{\small F}W_{40}^{60}=-2000\;\mathrm J \end{gather} \]

The total work done by the force F will be given by the sum of the three parts calculated above.

\[ \begin{gather} _{\small F}W_{0}^{60}=_{\small F}W_{0}^{25}+_{\small F}W_{25}^{40}+_{\small F}W_{40}^{60}\\[5pt] _{\small F}W_{0}^{60}=3000+0+(-2000) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {_{\small F}W_{0}^{60}=1000\;\mathrm J} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .