JEE Main 2019PhysicsLaws of MotionHardMCQ

JEE Main 2019Laws of Motion Question with Solution

JEE Main 2019 (12 Apr Shift 2)

Question

A block of mass 5 kg is (i) pushed in case A and (ii) pulled in case B, by a force F=20 N, making an angle of 30o with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is μ=0.2. The difference between the accelerations of the block, in case B and case A will be:
g=10m s-2

Choose an option

Show full solutionCorrect option: C
Correct answer
C0.8 m s-2

Step-by-step explanation

For Block A:

N=mg+Fsin30°
=5×10+20×12
=60 N
fr=0.260= 12 N
Fcos30° fr= ma
2032 -12=5×a
aA=103-125 ……(i)
For Block B:

Fsin30°+N=mg
N=5×10 20×12
=40 N
fr=0.2×40=8 N
Fcos30° fr=ma
2032-8=5×a
aB=103-85
aB-aA=45=0.8 m/s2

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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.