Question 67: Laws of Motion

Question

A block of mass $$5 \mathrm{~kg}$$ is placed on a rough inclined surface as shown in the figure. If $$\overrightarrow{F_1}$$ is the force required to just move the block up the inclined plane and $$\overrightarrow{F_2}$$ is the force required to just prevent the block from sliding down, then the value of $$\left|\overrightarrow{F_1}\right|-\left|\overrightarrow{F_2}\right|$$ is : [Use $$\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right]$$

Accepted Answer

$$\begin{aligned}
\mathrm{f}_{\mathrm{K}} & =\mu \mathrm{mg} \cos 	heta \
& =0.1 	imes \frac{50 	imes \sqrt{3}}{2} \
& =2.5 \sqrt{3} \mathrm{~N}
\end{aligned}$$
$$\begin{aligned}
\mathrm{F}_1= & \mathrm{mg} \sin 	heta+\mathrm{f}_{\mathrm{K}} \
= & 25+2.5 \sqrt{3}
\end{aligned}$$
$$\begin{aligned}
\mathrm{F}_2= & \mathrm{mg} \sin 	heta-\mathrm{f}_{\mathrm{K}} \
& =25-2.5 \sqrt{3} \
	herefore \mathrm{F}_1 & -\mathrm{F}_2=5 \sqrt{3} \mathrm{~N}
\end{aligned}$$

Answer

$$5 \sqrt{3} N$$

Answer

$$\frac{5 \sqrt{3}}{2} N$$

Answer

$$10 \mathrm{~N}$$

Answer

$$25 \sqrt{3} N$$

Laws of Motion

Choose the correct answer: