JEE Main 2024PhysicsLaws of MotionHardMCQ

JEE Main 2024 — Laws of Motion Question with Solution

JEE Main 2024 (31 Jan Shift 2)

Question

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F→1 is the force required to just move the block up the inclined plane and F→2 is the force required to just prevent the block from sliding down, then the value of F→1−F→2 is: [Use g=10 m s-2]

Choose an option

Show full solutionCorrect option: B
Correct answer
B53 N

Step-by-step explanation

Limiting friction force is, fmax=μmg cosθ

=0.1×50×32

=2.53 N

For first case:

We can write, F1=mg sinθ+fmax

=25+2.53

For second case:

F2=mg sinθ−fmax

=25−2.53

∴F1−F2=53 N

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About this question

This is a previous-year question from JEE Main 2024, covering the Laws of Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.