Question

An object with mass 7 kgmoves from a location $<22,43,-41>$ near the Earth's surface to location$<-27,11,46>\mathit{m}$. What is the change in the potential energy of the system consisting of the object plus the Earth?

Short answer

The change in potential energy of the system is $-2200\mathrm{J}$.

Step-by-step solution

Identification of the given data

The given data is listed as follows,

• The mass of the object is,m=7 kg
• The initial location is,
• The final location is,

Expression of change in potential energy and gravitational force

The change in potential energy is expressed as,

$\mathbf{∆}\mathit{U}\mathbf{=}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\left({\stackrel{\rightharpoonup }{r}}_f-{\stackrel{\rightharpoonup }{r}}_i\right)$

Here,${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}}_{\mathbf{i}}$is the final position,${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}}_{\mathit{f}}$ is the initial position and$\stackrel{\mathbf{\rightharpoonup }}{F}$is force.

The gravitational force acting on the object is shown as,

F=mg

Here, m is the mass, g is the acceleration due to gravity with value $9.8m/s^2$.

Determination of the value of the potential energy of the system

The expression for the change in potential energy is expressed as follows,

$∆U=mg\left({\stackrel{\rightharpoonup }{r}}_f-{\stackrel{\rightharpoonup }{r}}_i\right)$

It is known that the change of potential energy is in thedirection only.

Substitute all the values in the above expression.

$$\begin{gathered} ∆U=\left(7\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(11\mathrm{m}-43\mathrm{m}\right) \\ =\left(7\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(32\mathrm{m}\right) \\ =-2195.2\mathrm{kg}-{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}\times {\mathrm{m}}^2/{\mathrm{s}}^2} \\ =-2195.2\mathrm{J} \\ \approx -2200\mathrm{J} \end{gathered}$$

Thus, the change in potential energy of the system $-2200\mathrm{J}$.