First slide

Newton's laws of motion

Question

Five persons A, B, C, D and E are pulling a cart of mass 100 kg on a smooth surface and cart is moving with acceleration 3 m/s² in east direction. When person A stops pulling, it moves with acceleration 1 m/s² in the west direction. When person B stops pulling, it moves with acceleration 24 m/s² 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

Moderate

A

$26 m / s^{2}$
B

$3 \sqrt{71} m / s^{2}$
C

$25 m / s^{2}$
D

$30 m / s^{2}$

Solution

When all are pulling
${\bar{F}}_{n e t} = 100 \times 3 \hat{i} . . . . . . . (1)$
when A stops
${\bar{F}}_{n e t} - {\bar{F}}_{A} = 100 \times 1 (- \hat{i}) . . . . . . . (2)$
when B stops
${\bar{F}}_{n e t} - {\bar{F}}_{B} = 100 \times 24 \hat{j} . . . . . . . (3)$
from these three get ${\bar{F}}_{A} + {\bar{F}}_{B} = (700 \hat{i} - 2400 \hat{j}) N$
hence acceleration of the cart
$\begin{array}{l} \vec{a} = \frac{{\bar{F}}_{A} + {\bar{F}}_{B}}{m} = \frac{(700 \hat{i} - 2400 \hat{j})}{100} m / s^{2} \\ \vec{a} = (7 \hat{i} - 24 \hat{j}) m / s^{2} \Rightarrow | \vec{a} | = \sqrt{7^{2} + 24^{2}} = 25 m / s^{2} \end{array}$

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