The three spheres in Fig. 13-45, with masses${\mathbf{\text{m}}}_{\mathbf{\text{A}}}\mathbf{=}\mathbf{80}\mathit{k}\mathit{g}$, ${\mathbf{\text{m}}}_{\mathbf{\text{B}}}\mathbf{=}\mathbf{10}\mathit{g}$, and ${\mathbf{\text{m}}}_{\mathbf{\text{C}}}\mathbf{=}\mathbf{20}\mathit{g}$ , have their centers on a common line,with$\mathit{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{12}\mathit{c}\mathit{m}$ and $\mathit{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$ . You move sphereBalong the lineuntil its center-to-center separation fromCis $\mathit{d}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$ . Howmuch work is done on sphereB(a) by you and (b) by the net gravitational force onBdue to spheresAandC?

$$\begin{gathered} {\text{m}}_{\text{A}}=80g, \\ {\text{m}}_{\text{B}}=10\text{g,} \\ {\text{m}}_{\text{C}}=20\text{g} \\ \text{L}=12cm \\ \text{d}=4.0cm \end{gathered}$$

Using the formula for potential energy, find the potential energy of the system. By using the formula for work done, find thework done on sphere B by us and by the net gravitational force on due to Sphere Aand C.

Formulae are as follows:

$$\begin{gathered} \mathit{U}\mathbf{=}\mathbf{-}\frac{\mathbf{G}{\mathbf{m}}_{\mathbf{A}}{\mathbf{m}}_{\mathbf{B}}}{\mathbf{r}} \\ \mathit{W}\mathbf{=}{\mathit{U}}_{\mathbf{f}}\mathbf{-}{\mathit{U}}_{\mathbf{i}} \end{gathered}$$

where $m_A, m_B$are masses, r is the radius, G is gravitational constant, W is work done and U is potential energy.

Now,

$U=-\frac{G{m}_A{m}_B}{r}$

Therefore, initial potential energy is,

$U_i=-\frac{G{m}_A{m}_B}{d}-\frac{G{m}_A{m}_C}{L}-\frac{G{m}_B{m}_C}{L-d}$

The final potential energy is,

$U_f=\frac{G{m}_A{m}_B}{L-d}-\frac{G{m}_A{m}_C}{L}-\frac{G{m}_B{m}_C}{d}$

Hence, work done can be found as,

$$\begin{gathered} W=U_f-U_i \\ W=\left(\frac{G{m}_A{m}_B}{L-d}-\frac{G{m}_A{m}_C}{L}-\frac{G{m}_B{m}_C}{d}\right)-\left(-\frac{G{m}_A{m}_B}{d}-\frac{G{m}_A{m}_C}{L}-\frac{G{m}_B{m}_C}{L-d}\right) \\ W=G{m}_B\left[m_A\left(\frac{1}{d}-\frac{1}{L-d}\right)+m_C\left(\frac{1}{L-d}-\frac{1}{d}\right)\right] \\ W=G{m}_B\left[m_A\frac{L-2d}{d\left(L-d\right)}+m_C\frac{2d-L}{d\left(L-d\right)}\right] \\ W=G{m}_B\left(m_A-m_C\right)\frac{L-2d}{d\left(L-d\right)} \\ W\text{=}\left({\text{6.67×10}}^{\text{- 11}}\frac{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{2}}·\text{kg}}\right)\left(0.010Kg\right)\left(0.080kg-\text{0.020kg}\right)\left(\frac{0.12m-\text{2}\left(\text{0.040}\text{m}\right)}{\left(\text{0.040}\text{m}\right)\left(\text{0.12}-\text{0.040}\text{m}\right)}\right) \\ W=+5.0\times {10}^{-13}\text{J} \end{gathered}$$

Hence,the work done on sphere B by you is $+5.0\times {10}^{-13}\text{J}$.

Now,

$W=U_f-U_i$

Therefore, work done by gravity is,

$$\begin{gathered} W=-\left(U_f-U_i\right) \\ W=-5.0\times {10}^{-13}\text{J} \end{gathered}$$

Hence, the work done on the sphere by the net gravitational force on B due to Sphere A and C is$-5.0\times {10}^{-13}\text{J}$ .

Therefore, using the formula for potential energy, the work done can be found.