Question

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

Answer 1

Answer: 25

When all are pulling
${\bar{F}}_{\text{net}}=100\times 3\hat{i}\dots \left(1\right)$
when A stops
${\bar{F}}_{\text{net}}-{\bar{F}}_A=100\times 1\left(-\hat{i}\right)\dots \left(2\right)$
when B stops
${\bar{F}}_{net}-{\bar{F}}_B=100\times 24\hat{j}\dots \left(3\right)$
from these three get ${\bar{F}}_A+{\bar{F}}_B=\left(700\hat{i}-2400\hat{j}\right)N$
hence acceleration of the cart
$\vec{a}=\frac{{\bar{F}}_A+{\bar{F}}_B}{m}=\frac{\left(700\hat{i}-2400\hat{j}\right)}{100}m/s^2$
$\vec{a}=\frac{{\bar{F}}_A+{\bar{F}}_B}{m}=\frac{\left(700\hat{i}-2400\hat{j}\right)}{100}m/s^2$
$\vec{a}=\left(7\hat{i}-24\hat{j}\right)m/s^2$
$\Rightarrow \left|\vec{a}\right|=\sqrt{7^2+24^2}=25m/s^2$