Question

17. Three forces F1= (2i +4j) N; F2 =(2j -k)Nand F3= (k-4i-2$\hat{j}$ )Nare applied on an object

of mass 1 kg at rest at origin. The position of the object at t =2s will be:

(1) (-2 m, -6 m)

(2) (-4 m, 8 m)

(3) (3 m, 6 m)

(4) (2 m, -3 m)

Answer 1

Given Data:

$F_1=$ $\text{(2i +4j)}$ , $F_2=2\hat{j}-\hat{k}$, $F_3=\widehat{k-4\hat{i}}-2\hat{j}$

Mass of the object, $m=1kg$

Step 1: position of the object at $t=2sec$

Net force on an object, $\text{(2i +4j)}$ +$F_2=2\hat{j}-\hat{k}$ +$F_3=\widehat{k-4\hat{i}}-2\hat{j}$

$F_{net}=-2\widehat{i+4}\hat{j}$

Force on X-axis,$=2N$ Force on Y-axis, $=4N$

So, acceleration on X-axis, $a_x=\frac{ForceonX-axis}{mass}=\frac{-2}{1}=-2m{s}^{-2}$

So, acceleration on Y-axis, $a_y=\frac{ForceonY-axis}{mass}=\frac{4}{1}=4m{s}^{-2}$

Particle starts from rest at origin;

So, displacement in X-axis will be;

$S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2⟹S_x=\left(0\right)t+\frac{1}{2}(-2){\left(2\right)}^2=-4m$

displacement in Y-axis will be;

$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2⟹S_y=\left(0\right)t+\frac{1}{2}\left(4\right){\left(2\right)}^2=8m$

So, position of particle after, t=2sec is: $(-4m,8m)$