Force and trajectory

Data

• Mass m = 4 kg
• Equation of position: $\overrightarrow{r}=t^2\overrightarrow{i}-5 t \overrightarrow{j} - 2t^2 · \overrightarrow{k}$ m
• F?

Resolution

According to the Newton´s second law:

$${\sum }_{}\overrightarrow{F} = m·\overrightarrow{a}$$

We know the mass and know that the instantaneous acceleration is the derivative  of the instantaneous velocity, regarding instantaneous time and velocity, is the derivative of the vector of position regarding time.

$$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt} ⋮ \overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$$

Deriving the first vector of position and then the velocity vector:

$$\begin{gathered} \overrightarrow{v}=\frac{d\left(t^2·\overrightarrow{i}-5t·\overrightarrow{j}+2t^2·\overrightarrow{k}\right)}{dt}=2·t ·\overrightarrow{i} - 5 · \overrightarrow{j} - 4·t · \overrightarrow{k} \\ \overrightarrow{a}=\frac{d\left(2t ·\overrightarrow{i}-5\overrightarrow{j}-4t · \overrightarrow{k}\right)}{dt} = \underline{2·\overrightarrow{i} - 4· \overrightarrow{k} } \end{gathered}$$

Replacing the expression of the second law:

$$\begin{gathered} {\sum }_{}\overrightarrow{F}=4 kg·\left(2·\overrightarrow{i} - 4· \overrightarrow{k} \right) m/s\Rightarrow \\ \boxed{{\sum }_{}\overrightarrow{F}=8·\overrightarrow{i} - 16· \overrightarrow{k} N} \end{gathered}$$