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JEE MAIN 2024
310124 S2
Question
A block of mass 5" " kg is placed on a rough inclined surface as shown in the figure. If $\vec{F}_1$ is the force required to just move the block up the inclined plane and $\overrightarrow{\mathrm{F}}_2$ is the force required to just prevent the block from sliding down, then the value of $\left|\vec{F}_1\right|-\left|\vec{F}_2\right|$ is : [Use $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
Select the correct option:
A
$25 \sqrt{3} \mathrm{~N}$
B
$50 \sqrt{3} \mathrm{~N}$
C
$\frac{5 \sqrt{3}}{2} \mathrm{~N}$
D
10 N
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} \mathrm{f}_{\mathrm{K}} & =\mu \operatorname{mgcos} \theta \\ & =0.1 \times \frac{50 \times \sqrt{3}}{2} \\ & =2.5 \sqrt{3} \mathrm{~N}\end{aligned}$ $\begin{gathered}\mathrm{F}_2=\mathrm{mgsin} \theta-\mathrm{f}_{\mathrm{K}} \\ \quad=25-2.5 \sqrt{3} \\ \therefore \mathrm{~F}_1-\mathrm{F}_2=5 \sqrt{3} \mathrm{~N}\end{gathered}$
Question Tags
JEE Main
Physics
Medium
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