JEE Main 2019PhysicsLaws of MotionMediumMCQ

JEE Main 2019 — Laws of Motion Question with Solution

JEE Main 2019 (09 Jan Shift 1)

Question

A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block does not move downward? (take g=10 ms-2 )

Choose an option

Show full solutionCorrect option: D
Correct answer
D32 N

Step-by-step explanation



Block will be at rest

mgsin⁡θ+3N=P+f

10×10×12+3=P+μmgcos⁡θ

( ∵ To get P minimum friction must be minimum)

Pmin=502+3-0.6×10×10×12

=202+3=32N

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About this question

This is a previous-year question from JEE Main 2019, 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.