JEE Main 2026 — Laws of Motion Question with Solution
JEE Main 2026 (28 January Shift 1)
Question
A block of mass 5 kg is moving on an inclined plane which makes an angle of with the horizontal. Friction coefficient between the block and inclined plane surface is . The force to be applied on the block so that the block will move down without acceleration is N.
.
.
Choose an option
Show full solutionCorrect option: C
Correct answer
C12.5
Step-by-step explanation
For motion without acceleration, net force = 0.
Component of weight along incline: N (down).
Normal force: N.
Friction force opposing downward motion:
N (up).
For equilibrium along the incline, applying force F down the incline:
, so , giving N.
Component of weight along incline: N (down).
Normal force: N.
Friction force opposing downward motion:
N (up).
For equilibrium along the incline, applying force F down the incline:
, so , giving N.
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Laws of Motion chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2026, 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.