Question

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?

Short answer

The speed of the object derived from the given measurements on its final location is 10.15 m/s .

Step-by-step solution

Given measurements

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 $\left(F_{2x},F_{2y},F_{2z}\right)=\left(140,250,150\right)m$

Concept of the energy principle 

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, $K=\frac{1}{2}m{v}^2\left(m{c}^2\right)$

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

Determine the value displacement

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.

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

Determine the value of the kinetic energy

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

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