Question

Five persons $A,B,C,D$ and $E$ are pulling a cart of mass $100\text{kg}$ on a smooth surface and cart is moving with acceleration $3{\text{m/s}}^2$ in east direction. When person $A$ stops pulling, it moves with acceleration $1{\text{m/s}}^2$ in the west direction. When person $B$ stops pulling, it moves with acceleration $24{\text{m/s}}^2$ in the north direction. The magnitude of acceleration of the cart when only $A$ and $B$ pull the cart keeping their directions same as the old directions, is

• A. $24{\text{m/s}}^2$
• B. $3\sqrt{71}{\text{m/s}}^2$
• C. $30{\text{m/s}}^2$
• D. $25{\text{m/s}}^2$

Answer 1

The correct option is D $25{\text{m/s}}^2$

$\begin{array}{}{\overrightarrow{F}}_A+{\overrightarrow{F}}_B+{\overrightarrow{F}}_C+{\overrightarrow{F}}_D+{\overrightarrow{F}}_E=(100\times 3)\hat{i}& \text{(1)}\end{array}$

If $A$ stops pulling

${\overrightarrow{F}}_B+{\overrightarrow{F}}_C+{\overrightarrow{F}}_D+{\overrightarrow{F}}_E=(100\times 1)\widehat{(}-i)$
By putting this in equation 1 we get
$\Rightarrow {\overrightarrow{F}}_A=300\hat{i}+100\hat{i}$
$\Rightarrow {\overrightarrow{F}}_A=400\hat{i}\text{N}$

If $B$ stops pulling
${\overrightarrow{F}}_A+{\overrightarrow{F}}_C+{\overrightarrow{F}}_D+{\overrightarrow{F}}_E=(100\times 24)\hat{j}$
Putting in equation 1 we get
$\Rightarrow {\overrightarrow{F}}_B=300\hat{i}-2400\hat{j}$
${\overrightarrow{F}}_B=100.(3\hat{i}-24\hat{j})$

Now, ${\overrightarrow{F}}_A+{\overrightarrow{F}}_B=(7\hat{i}-24\hat{j}).100$

Hence, if only $A$ and $B$ are pulling,
${\overrightarrow{F}}_A+{\overrightarrow{F}}_B=100.a_{\text{net}}$
$100.(7\hat{i}-24\hat{j})=100a_{\text{net}}$
$a_{\text{net}}=7i-24j$
$|a_{\text{net}}|=25{\text{m/s}}^2$