Three forces ${\to F}_{1} = (2^i + 4^j) N; {\to F}_{2} = (2^j -^k) N$ and ${\to F}_{3} = (^k - 4^i - 2^j) N$ are applied on an object of mass $1 kg$ at rest at origin. The position of the object at $t = 2 s$ will be

(- 2 m, - 6 m)

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(- 4 m, 8 m

)

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(3 m, 6 m)

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(2 m, - 3 m)

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Solution

The correct option is B $(- 4 m, 8 m$ )
From Newton's second law,
$F_{net} = {\to F}_{1} + {\to F}_{2} + {\to F}_{3} = m {\to a}_{net}$
$\Rightarrow 2 i + 4 j + 2 j - k + k - 4 i - 2 j = 1 \to a$
$\Rightarrow \to a = - 2 i + 4 j$
$\Rightarrow$ Now $\to s = \to u t + \frac{1}{2} \to a t^{2}$
as body is initially at rest
$\to s = \frac{1}{2} \to a t^{2}$
$\to s = \frac{1}{2} (- 2 i + 4 j) t^{2}$
$\to s = - 4 i + 8 j m$
So $(- 4 m, 8 m)$

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Three forces →F1=(2^i+4^j)N; →F2=(2^j−^k)N and →F3=(^k−4^i−2^j)N are applied on an object of mass 1 kg at rest at origin. The position of the object at t=2s will be

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