Question

A body of mass $5kg$ under the action of constant force $\overrightarrow{F}=F_X\hat{i}+F_Y\hat{j}$ has velocity at $t=0$ as $\overrightarrow{v}=(6\hat{i}-2\hat{j})m/s$ and at $t=10s$ as $\overrightarrow{v}=+6\hat{j}m/s$. The force $\overrightarrow{F}$ is then

• A. $(\frac{3}{5}\hat{i}-\frac{4}{5}\hat{j})N$
• B. $(-3\hat{i}+4\hat{j})N$
• C. $(3\hat{i}+4\hat{j})N$
• D. $-(\frac{3}{5}\hat{i}+\frac{4}{5}\hat{j})N$

Answer 1

Answer: (B)

$(-3\hat{i}+4\hat{j})N$

Given :

$\overrightarrow{F}=Fx\hat{i}+Fy\hat{j}$

t=0 $\overrightarrow{v}=6\hat{i}-2\hat{j}$

t=10s $\overrightarrow{v}=+6\hat{j}$

Solution :

Case 1 (x component)

$v_x=u_x+a_{x}t$

$0=6+10a_x$

$a_x=\frac{-3}{5}$

$Fx=m{a}_x$ = -3

Case 2 (y component)

$v_y=u_y+a_{y}t$

$6=(-2)+10a_y$

$a_y=\frac{4}{5}$

$Fy=m{a}_y$ = 4

$\overrightarrow{F}=Fx\hat{i}+Fy\hat{j}$

$\overrightarrow{F}=-3\hat{i}+4\hat{j}$

The Correct Opt = B