An object with mass 100kg moved in outer space. When it was at location $\left(9,-24,4\right)$its speed was $3.5m/s$. A single constant

force $\left(250,400,-170\right)W$acted on the object while the object moved from location $\left(9,-24,-4\right)$to location $\left(15,-17,-8\right)$. Then a different single constant force $\left(140,250,150\right)$N acted on the object while the object moved from location $\left(15,-17,-8\right)$ to location $\left(9,-24,-4\right)$. What is the speed of the object at this final location?

Mass of the object is $m=100kg$

Initial speed is $v_i=3.5m/s$

Initial position of the object $\left(x_1,y_1,z_1\right)=\left(9,-24,-4\right)$

Position of the object at second stage $\left(x_2,y_2,z_3\right)=\left(15,-17,-8\right)$

Position of the object at the final stage $\left(x_3,y_3,z_3\right)=\left(19,-24,-3\right)$

Force exerted on the object at the end of the second stage $\left(F_{1x},F_{1y},F_{1z}\right)=\left(250,400,-170\right)m$

Force exerted on the object at the end of the third stage $\left(F_{2x},F_{2y},F_{2z}\right)=\left(140,250,150\right)m$

The change in energy of a system or object is equal to the work done and energy input of the surroundings.

This can be expressed as, ${\mathit{E}}_{\mathbf{f}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{E}}_{\mathbf{i}}\mathbf{+}\mathit{W}$

Which can be further expressed as, $\left(K_f+m{c}^2\right)=\left(K_i+m{c}^2\right)+W$

The formula above can be broken down into another expression where in the rest energy $\left(m{c}^2\right)$ can be cancelled out since there is no change in its rest energy. In this expression, $K=\frac{1}{2}m{v}^2\left(m{c}^2\right)$

$\begin{array}{l}\left(\frac{1}{2}m{v}_f^2+m{c}^2\right)=\left(\frac{1}{2}m{v}_f^2+m{c}^2\right)+w\\ \Rightarrow \frac{1}{2}m{v}_f^2=\frac{1}{2}m{v}_i^2+W\end{array}$

Substitute the values then solve for the displacement at the end of the second stage using the first formula above.

Substitute the values then solve for the displacement at the end of the third stage using the same method.

$\begin{array}{l}\overrightarrow{d_2}=\left(x_3,y_3,z_3\right)-\left(x_2,y_2,z_2\right)\\ \overrightarrow{d_2}==(19,-24,-3\}-\left(15-17,-8\right)\\ \overrightarrow{d_2}=(4,-7,5)m\end{array}$

$$\begin{gathered} E_f=E_i+W \\ \frac{1}{2}m{v}_f^2=\frac{1}{2}m{v}_i^2+\left(F_1.\overrightarrow{d_1}\right)+\left(F_2.\overrightarrow{d_2}\right) \\ \frac{1}{2}\left(100Kg\right)(v_f^2-{\left(3.5m/s\right)}^2=\left[(\left(250,400,-170\right)N-\left(6,7,-4\right)m)\left(\right(140,250,150)N-(4,-7,5)m\right] \\ 50v_f^2-612.5J=4540J \\ v=\sqrt{\frac{4540+612.5}{50}} \end{gathered}$$

Therefore, the speed of the object on its final location is 10.15 m/s .