JEE MAIN - Physics (2024 - 31st January Evening Shift - No. 6)
_31st_January_Evening_Shift_en_6_1.png)
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 $$\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right]$$
Explanation
$$\begin{aligned} \mathrm{f}_{\mathrm{K}} & =\mu \mathrm{mg} \cos \theta \\ & =0.1 \times \frac{50 \times \sqrt{3}}{2} \\ & =2.5 \sqrt{3} \mathrm{~N} \end{aligned}$$
_31st_January_Evening_Shift_en_6_2.png)
$$\begin{aligned} \mathrm{F}_1= & \mathrm{mg} \sin \theta+\mathrm{f}_{\mathrm{K}} \\ = & 25+2.5 \sqrt{3} \end{aligned}$$
_31st_January_Evening_Shift_en_6_3.png)
$$\begin{aligned} \mathrm{F}_2= & \mathrm{mg} \sin \theta-\mathrm{f}_{\mathrm{K}} \\ & =25-2.5 \sqrt{3} \\ \therefore \mathrm{F}_1 & -\mathrm{F}_2=5 \sqrt{3} \mathrm{~N} \end{aligned}$$
Comments (0)
