91Ó°ÊÓ

The mass of an object is,$m=250\mathrm{kg}$.

The position vector of the object, $\stackrel{→}{r}=\left(3500-160t\right)\hat{i}+2700\hat{j}+300\hat{k}$.

Here,we can apply the calculus knowledge along with Newton’s second law of motion.

Formula:

$$\begin{gathered} \stackrel{→}{v}=\frac{d\stackrel{→}{r}}{dt} \\ \stackrel{→}{P}=m\stackrel{→}{v} \end{gathered}$$

The equation for velocity is,

$\stackrel{→}{v}=\frac{d\stackrel{→}{r}}{dt}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \stackrel{→}{v}=\frac{d\left(\left(3500-160t\right)\hat{i}+2700\hat{j}+300\hat{k}\right)}{dt} \\ =-160\hat{i} \end{gathered}$$

Linear momentum can be calculated as,

$\stackrel{→}{P}=m\stackrel{→}{v}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \stackrel{→}{P}=250×\left(-160\hat{i}\right) \\ =\left(-4.0×{10}^4\mathrm{kg}.\frac{\mathrm{m}}{\mathrm{s}}\right) \end{gathered}$$

Thus, the linear momentum of an object, $\stackrel{→}{P}=\left(-4.0×{10}^4\mathrm{kg}.\frac{\mathrm{m}}{\mathrm{s}}\right)$.

From the equation of velocity, we can say that the direction of motion of the object is (- x) direction.

Thus, the direction of motion is in the -x (west) direction .

The equation of force in terms of momentum is,

$\stackrel{→}{F}=\frac{d\stackrel{→}{P}}{dt}$

From part a, since the value of P does not change with time, we get,

$\stackrel{→}{F}=0$

Thus, the net force on the object,$\stackrel{→}{F}=0$