91Ó°ÊÓ

1. m = 12 kg
2. $V_x=5m/s$
   
3. $F_y=17N$
   
4. $t=3sec$
   

Acceleration of an object produced by a net force is directly proportional to the product of mass and acceleration is known as Newton’s second law of motion. So to calculate net force use the mass of the object and acceleration of the object.

Formula

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{m}\mathbf{×}\mathbf{a}$

Here,${\mathit{F}}_{\mathbf{net}}$ is the force, m is the mass of the object, and a is the acceleration of the object.

(a)

To find the velocity vector, first, find velocity components along x and y directions.

We have,

$V_x=5\hat{i}$

To find y component of velocity, first calculate acceleration in y direction:

role="math" localid="1660910282733" $$\begin{gathered} F_y=m×a_y \\ 17=a_y×12 \\ {a}_y=1.41m/s \end{gathered}$$

Now velocity along direction,

$\begin{array}{l}{v}_y=v_0+a_y×t\\ v_y=1.41×3\\ =4.23\mathrm{m}/\mathrm{s}\end{array}$

So, velocity vector,

$$\begin{gathered} \stackrel{r}{v}=v_x\hat{i}+y_y\hat{j} \\ \stackrel{r}{v}=5\hat{i}+4.3\hat{j} \end{gathered}$$

Therefore, the position vector is $5\hat{i}+4.3\hat{j}$

(b)

Position along x axis:

$r_x=v_x×t=5×3=15m$

Position along y axis:

$\begin{array}{l}{r}_y=V_{0}t+\frac{1}{2}{a}_y{t}^2\\ =\frac{1}{2}×1.41×3^2\\ =6.345m\end{array}$

So, position vector:

$\begin{array}{l}r=r_x\stackrel{Áåœ}{i}+r_y\hat{j}\\ =15\hat{i}+6.4\hat{j}\end{array}$

Therefore, the position vector is $15\hat{i}+6.4\hat{j}$.