Question

Figure 8-20 shows one direct path and four indirect paths from point i to point f. Along the direct path and three of the indirect paths, only a conservative force${\mathit{F}}_{\mathbf{c}}$acts on a certain object. Along the fourth indirect path, both${\mathit{F}}_{\mathbf{c}}$and a nonconservative force${\mathit{F}}_{\mathbf{n}\mathbf{c}}$act on the object. The change $\mathbf{∆}{\mathit{E}}_{\mathbf{m}\mathbf{e}\mathbf{c}}$in the object’s mechanical energy (in joules) in going from i to f is indicated along each straight-line segment of the indirect paths. What is$\mathbf{∆}{\mathit{E}}_{\mathbf{mec}}$(a) from i to f along the direct path and (b) due to ${\mathit{F}}_{\mathbf{n}\mathbf{c}}$along the one path where it acts?

Short answer

a) The $∆{\mathrm{E}}_{\mathrm{mec}}$ from $i$to $f$along the direct path is 12 J .

b) The $∆{\mathrm{E}}_{\mathrm{mec}}$due to $F_{nc}$ along the one path where it acts is -2 J .

Step-by-step solution

Given information 

A figure which shows one direct path and four indirect paths from point $i$to$f$

To understand the concept

The problem involves the conservative force which is a force donein moving a particle from one point to another, such that the force is independent of the path taken by the particle.Here the concept of change in mechanical energy due to a conservative force which is independent of the path can be used. From this, the $\mathbf{∆}{\mathit{E}}_{\mathbf{mec}}$from $\mathit{i}$to $\mathit{f}$along the direct path and due to ${\mathit{F}}_{\mathbf{n}\mathbf{c}}$along one path where it acts can be found.

Formula:

Mechanical energy is given by,

$∆E_{mec}=\sum _{}E$

(a) To find ∆Emec from i to f along the direct path

$$\begin{gathered} ∆E_{mec}=\sum _{}E \\ \mathrm{Therefore},\mathrm{change}\mathrm{in}∆{\mathrm{E}}_{\mathrm{mec}}\mathrm{along}\mathrm{each}\mathrm{path}\mathrm{is} \\ ∆{\mathrm{E}}_{\mathrm{mec}}=32-32+10 \\ =12\mathrm{J} \\ ∆{\mathrm{E}}_{{\mathrm{mec}}_2}=20-6-4 \\ =10\mathrm{J} \\ ∆{\mathrm{E}}_{{\mathrm{mec}}_3}=\mathrm{not}\mathrm{given} \\ ∆{\mathrm{E}}_{{\mathrm{mec}}_3}=15+7-10 \\ =12\mathrm{J} \\ ∆{\mathrm{E}}_{{\mathrm{mec}}_4}=40-30+2 \\ =12\mathrm{J} \end{gathered}$$

As the $F_{nc}$does not acts along the direct path, the role="math" localid="1657186250637" $∆E_{mec}$from $i$to $f$along the direct path is 12 J

(b) To find ∆Emec due to Fnc along the one path where it acts

We have$∆E_{mec}$for path 2,

$$\begin{gathered} ∆E_{me{c}_2}=20-6-4 \\ =10J \end{gathered}$$

Therefore,$∆E_{mec}$due to non-conservative force is,

$$\begin{gathered} ∆{\mathrm{E}}_{\mathrm{mec}}=∆{\mathrm{E}}_{{\mathrm{mec}}_2}-∆{\mathrm{E}}_{{\mathrm{mec}}_1} \\ ∆{\mathrm{E}}_{\mathrm{mec}}=10\mathrm{J}-12\mathrm{J} \\ ∆{\mathrm{E}}_{\mathrm{mec}}=-2\mathrm{J} \end{gathered}$$

Therefore, the $∆E_{mec}$ due to $F_{nc}$along the one path where is acts is -2 J