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A $0.2 \mathrm{~kg}$ object at rest is subjected to a force $(0.3 \hat{i}-0.4 \hat{j}) \mathrm{N}$. What is the velocity after $6 \mathrm{~s}$ ?
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Verified Answer
The correct answer is:
$(9 \hat{i}-12 \hat{j})$
Here, $m=0.2 \mathrm{~kg}, u=0$
$$
\begin{aligned}
& \vec{F}=(0.3 \hat{i}-0.4 \hat{j}) \\
& \vec{v}=?, t=6 \mathrm{~s} \\
& \vec{a}=\frac{\vec{F}}{m}=\frac{(0.3 \hat{i}-0.4 \hat{j})}{0.2}=\left(\frac{3}{2} \hat{i}-2 \hat{j}\right)
\end{aligned}
$$
From $\vec{v}=\vec{u}+\vec{a} t$
$$
\vec{v}=0+\left(\frac{3}{2} \hat{i}-2 \hat{j}\right) \times 6=9 \hat{i}-12 \hat{j}
$$
$$
\begin{aligned}
& \vec{F}=(0.3 \hat{i}-0.4 \hat{j}) \\
& \vec{v}=?, t=6 \mathrm{~s} \\
& \vec{a}=\frac{\vec{F}}{m}=\frac{(0.3 \hat{i}-0.4 \hat{j})}{0.2}=\left(\frac{3}{2} \hat{i}-2 \hat{j}\right)
\end{aligned}
$$
From $\vec{v}=\vec{u}+\vec{a} t$
$$
\vec{v}=0+\left(\frac{3}{2} \hat{i}-2 \hat{j}\right) \times 6=9 \hat{i}-12 \hat{j}
$$
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