Question 206: Laws of Motion

Question

A body of mass $$2 \mathrm{~kg}$$ is acted upon by two forces each of magnitude $$1 \mathrm{~N}$$ and inclined at $$60^{\circ}$$ with each other. The acceleration of the body in $$\mathrm{m} / \mathrm{s}$$ is [$$\cos 60^{\circ}=0.5$$]

Accepted Answer

Given, $$F_1=F_2=1 \mathrm{~N}, m=2 \mathrm{~kg}$$ and $$	heta=60^{\circ}$$
Resultant force,
$$\begin{aligned}
F & =\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos 	heta} \
& =\sqrt{1^2+1^2+2 \cos 60^{\circ}} \
& =\sqrt{2+2(0.5)}=\sqrt{3} \mathrm{~N}
\end{aligned}$$
$$\begin{array}{ll}
\Rightarrow & m a=\sqrt{3} \
	ext { or } & a=\frac{\sqrt{3}}{m}=\frac{\sqrt{3}}{2}=\sqrt{\frac{3}{4}}=\sqrt{0.75} \mathrm{~ms}^{-2}
\end{array}$$

Answer

$$\sqrt{0.75}$$

Answer

$$\sqrt{0.65}$$

Answer

$$\sqrt{0.20}$$

Answer

$$\sqrt{0.35}$$

Laws of Motion

A body of mass $2 \mathrm{~kg} is acted upon by two forces each of magnitude 1 \mathrm{~N} and inclined at 60^{\circ} with each other. The acceleration of the body in \mathrm{m} / \mathrm{s} is [\cos 60^{\circ}=0.5$]

Choose the correct answer: