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JEE Main 2021 (1 Sep, Shift 2) · Inertia, First & Second Laws

Inertia, First & Second Laws question from JEE Main 2021 (1 Sep, Shift 2)

Q21JEE Main 2021 · 1 SEP · SHIFT 2Medium

Inertia, First & Second Laws

An object of mass 'm' is being moved with a constant velocity under the action of an applied force of 2N along a frictionless surface with following surface profile.

Figure

The correct applied force vs distance graph will be:

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Answer: (c)

Fmg sinθFvup: F along v (+2 N)down: F against v (−2 N)
  1. Constant velocity on a frictionless incline: applied force balances mgsin⁡θmg\sin\theta, so it always points up the slope with magnitude 2 N.
  2. Going up the left face (first D/2D/2), the force is along the motion: F=+2 NF = +2\text{ N}.
  3. Going down the right face (next D/2D/2), the force still points up the slope, i.e. against the motion: F=−2 NF = -2\text{ N}.
  4. So the graph is a +2 N step then a −2 N step. The answer is (c).
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