Question

Outside the space shuttle, you and a friend pull on two ropes to dock a satellite whose mass is $\mathbf{700}\mathbf{}\mathbf{\text{kg}}$. The satellite is initially at position$\mathbf{(}\mathbf{3}\mathbf{.5}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{.4}\mathbf{)}\mathbf{\text{m}}$and has a speed of$\mathbf{4}\mathbf{}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$. You exert a force$\mathbf{(}\mathbf{-}\mathbf{400}\mathbf{,}\mathbf{310}\mathbf{,}\mathbf{-}\mathbf{250}\mathbf{)}\mathbf{\text{N}}$. When the satellite reaches the position$\mathbf{(}\mathbf{7}\mathbf{.1}\mathbf{,}\mathbf{3}\mathbf{.2}\mathbf{,}\mathbf{1}\mathbf{.2}\mathbf{)}\mathbf{\text{m}}$its speed is$\mathbf{4}\mathbf{.01}\mathbf{}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$. How much work did your friend do?

Short answer

The work done by your friend is$-133.9\text{J}$ , where the negative sign means that the satellite applies a work done on your friend.

Step-by-step solution

Given measurements

The satellite has mass $\mathrm{m}=700\text{\hspace{0.17em}kg}$, an initial speed${\mathrm{v}}_{\mathrm{i}}=4.0\text{\hspace{0.17em}m/s}$ , its final speed is ${\text{v}}_{\text{f}}=4.01\text{\hspace{0.17em}m/s}$.

Concept of the work done and kinetic energy

According to the principle of energy,

${\mathbf{E}}_{\mathbf{f}}\mathbf{=}{\mathbf{E}}_{\mathbf{i}}\mathbf{+}\mathbf{W}$

Where${\mathbf{E}}_{\mathbf{f}}$and${\mathbf{E}}_{\mathbf{i}}$are the final and initial energies of the satellite respectively.

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation will be in the form

$\mathbf{(}{\mathbf{K}}_{\mathbf{f}}\mathbf{+}{\mathbf{mc}}^{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{(}{\mathbf{K}}_{\mathbf{i}}\mathbf{+}{\mathbf{mc}}^{\mathbf{2}}\mathbf{)}\mathbf{+}{\mathbf{W}}_{\mathbf{1}}\mathbf{+}{\mathbf{W}}_{\mathbf{2}}$.

Determine the value of work done by you

${\mathrm{E}}_{\mathrm{f}}={\mathrm{E}}_{\mathrm{i}}+\mathrm{W}$Where${\text{E}}_{\text{f}}$and ${\text{E}}_{\text{i}}$are the final and initial energies of the satellite respectively.

Whileis the work done by the surrounding (you and your friend).

Let the work you done is${\mathrm{W}}_1$and the work done by your friend is${\mathrm{W}}_2$.

Hence, the total work done is${\mathrm{W}}_{\mathrm{t}}={\mathrm{W}}_1+{\mathrm{W}}_2$.

Also, the system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed.

The summation of both energies called the particle energy. So, equation will be in the form

$({\mathrm{K}}_{\mathrm{f}}+{\mathrm{mc}}^2)=({\mathrm{K}}_{\mathrm{i}}+{\mathrm{mc}}^2)+{\mathrm{W}}_1+{\mathrm{W}}_2$

Where is the mass of satellite, $\mathrm{c}$is the speed of light which equals $3\times {10}^8\text{m}/\text{s}$and the term ${\text{mc}}^2$represents the energy at rest and the term $\text{K}$is the kinetic energy. In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the satellite at low speeds where $\mathrm{K}=\frac{1}{2}{\text{mv}}^2$and equation will be in the form

role="math" localid="1657114002798" $\begin{array}{c}\left(\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2+{\mathrm{mc}}^2\right)=\left(\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2+{\mathrm{mc}}^2\right)+{\mathrm{W}}_1+{\mathrm{W}}_2\\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2=\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2+{\mathrm{W}}_1+{\mathrm{W}}_2\end{array}$

The work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Given the force you applied, therefore calculate the work done by you${\mathrm{W}}_1$ in the next form

${\mathrm{W}}_1=\mathrm{F}\cdot \mathrm{\Delta }$

Where$\mathrm{F}$ is the force applied by you and equals $(-400,310,-250)\text{\hspace{0.17em}N}$. The satellite moves from an initial position${\mathrm{r}}_{\mathrm{i}}=(3.5,-1,2.4)\text{\hspace{0.17em}m}$ to a final position${\mathrm{r}}_{\mathrm{f}}=(7.1,3.2,1.2)\text{\hspace{0.17em}m}$ , so the displacement $\mathrm{\Delta }\text{r}$of the satellite by you could be calculated by

$\begin{array}{c}\mathrm{\Delta r}={\mathrm{r}}_{\mathrm{f}}-{\mathrm{r}}_{\mathrm{i}}\\ =(7.1,3.2,1.2)\text{\hspace{0.17em}m}-(3.5,-1,2.4)\text{\hspace{0.17em}m}\\ =(3.6,4.2,-1.2)\text{\hspace{0.17em}m}\end{array}$

Now plug these values for$F$ and$∆$ r into equation to get the total work done on the satellite by you

$\begin{array}{c}{\mathrm{W}}_1=\mathrm{F}\cdot \mathrm{\Delta r}\\ =(-400,310,-250)\text{\hspace{0.17em}N}\times (3.6,4.2,-1.2)\text{\hspace{0.17em}m}\\ =162\text{\hspace{0.17em}J}\end{array}$

Determine the values to get the work done by your friend

Now plug these values for${\mathrm{v}}_{\mathrm{i}},{\mathrm{v}}_{\mathrm{f}},\mathrm{m}$ &${\mathrm{W}}_1$ to get${\mathrm{W}}_2$

$\begin{array}{c}{\mathrm{W}}_2=\frac{1}{2}\mathrm{m}[{\mathrm{v}}_{\mathrm{f}}^2-{\mathrm{v}}_{\mathrm{i}}^2]-{\mathrm{W}}_1\\ =\frac{1}{2}[700\text{kg}][{(4.01{\text{\hspace{0.17em}ms}}^{-1})}^2-{(4.0{\text{ms}}^{-1})}^2]-162\text{\hspace{0.17em}J}\\ =-133.9\text{\hspace{0.17em}J}\end{array}$

The work done by your friend is $-133.9\text{\hspace{0.17em}J}$, where the negative sign means that the satellite applies a work done on your friend.