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Physics · Newton's Laws of Motion

JEE Main 2026 — 5 April, Evening Shift — Question 3

A mass of 1kg1\mathrm{kg} kept in an inclined plane with 30∘30^{\circ} inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of 4m/s4\mathrm{m/s}. The work done by the frictional force in time 2s2\mathrm{s} is J.

  1. Option A:

    20

    Correct
  2. Option B:

    25

  3. Option C:

    30

  4. Option D:

    10

Answer: A

Step-by-step solution

Friction f=mgsin⁡30∘=5f = mg\sin30^\circ = 5 N. Distance moved in 2 s: d=vt=8d = vt = 8 m. Work done by friction: W=fdcos⁡60∘=5×8×0.5=20W = f d \cos60^\circ = 5\times8\times0.5 = 20 J.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Physics
Chapter
Newton's Laws of Motion
Topic
Free Body Diagrams and Constraint Relations
A mass of 1 kg kept in an inclined plane with 30 ° inclination with… | JEE Main 2026 PYQ with Solution · DhiX AI