91Ó°ÊÓ

The mass ofa box is$20kg$

The constant force is$(90,0,0)N$

At t = 5,the location of a box is$(8,2,-1)m$

The travelling velocity is$(3,0,0)m/s$

Time$t=5.6s$

$\mathit{F}\mathbf{}\mathbf{=}\mathbf{}\mathit{m}\mathit{a}$

Lets assume the components of force$(x,y,z)$

The force in X direction is $90N$and addition to there is friction also $⇒(90-f)$

The force in Y direction is$0N$

The force in Z direction is the force due to gravity force balanced by the normal force ⇒$(N-mg=0)$

Therefore,

$F=ma=(90-f,0,N-mg=0)$ …(1)

Where,

$$\begin{gathered} f=0.25×m×g \\ f=0.25×20×g \\ =5g \\ f=5×10 \\ f=50N \end{gathered}$$

Substitute in Equation (1),

$F=ma$

Where,

$F=(90-50,0,0)⇒(40,0,0)$ which means

$a=(2,0,0)$

We Know that,

$$\begin{gathered} \left(x,y,z\right)=\left(x_0,y_0,z_0\right)+\left[\left(v{x}_0,v{y}_0,v{z}_0\right)t\right]+\left[5\left(a{t}^2\right)\right] \\ =\left(8,2,-1\right)+\left[\left(3,0,0\right)t\right]+\left[5\left(2,0,0\right)t^2\right] \\ =\left(8+3t+t^2,2,-1\right) \end{gathered}$$

Put $t=5.6s$in $(x,y,z)⇒$

$$\begin{gathered} \left(x,y,z\right)=\left[\left(8+\left(3×5.6\right)+{\left(5.6\right)}^2\right),2,-1\right] \\ =\left(56.16,2,-1\right) \end{gathered}$$

Hence, the position of the box is$\left(x,y,z\right)\mathbf{=}\left(56.16,2,-1\right)$

To find Velocity,

The velocity is the derivative of $\left(8+3t+t^2,2,-1\right)⇒\left(3+2t,0,0\right)$

Put $t=5.6,$

$$\begin{gathered} velocity=(\left(3+2×5.6\right),0,0) \\ =\left(14.2,0,0\right)m/s \end{gathered}$$

Hence the velocity is$\mathbf{(}\mathbf{14}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{)}\mathit{m}\mathbf{/}\mathit{s}$