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A body of mass $2 \mathrm{~kg}$ is acted upon by two forces each of magnitude $1 \mathrm{~N}$, making an angle of $60^{\circ}$ with each other. The net acceleration of the body (in $\mathrm{ms}^{-2}$ ) is
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The correct answer is:
$\frac{\sqrt{3}}{2}$
Given that, mass of body, $m=2 \mathrm{~kg}$
Two forces $F_1$ and $F_2$ each of magnitude $1 \mathrm{~N}$
are acting on the body by making an angle $\theta=60^{\circ}$
with each other, then magnitude of resultant force by addition of two vectors is
Net force, $F=\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos 60^{\circ}}$
Substituting the values, we get
$F=\sqrt{(\mathrm{l})^2+(\mathrm{l})^2+2(\mathrm{l})(\mathrm{l}) \times \frac{\mathrm{l}}{2}}$
$=\sqrt{3} \mathrm{~N}$
By Newton's second law of motion,
$F=m a$
$\therefore$ Acceleration of body,
$a=\frac{F}{m}=\frac{\sqrt{3}}{2} \mathrm{~m} / \mathrm{s}^2$
Two forces $F_1$ and $F_2$ each of magnitude $1 \mathrm{~N}$
are acting on the body by making an angle $\theta=60^{\circ}$
with each other, then magnitude of resultant force by addition of two vectors is
Net force, $F=\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos 60^{\circ}}$
Substituting the values, we get
$F=\sqrt{(\mathrm{l})^2+(\mathrm{l})^2+2(\mathrm{l})(\mathrm{l}) \times \frac{\mathrm{l}}{2}}$
$=\sqrt{3} \mathrm{~N}$
By Newton's second law of motion,
$F=m a$
$\therefore$ Acceleration of body,
$a=\frac{F}{m}=\frac{\sqrt{3}}{2} \mathrm{~m} / \mathrm{s}^2$
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