Home/
Competitive Exams/
JEE Main & Advanced/
Physics/
Laws of Motion and Friction/
Question 30268
JEE Physics · Hard

Laws of Motion and Friction: JEE — Multiple Choice MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

Timer-based practiceDetailed answer reviewMobile-friendly flow
JEE — Multiple ChoiceHardQuestion 30268

Question

A block of mass 10 kg is on a rough surface (μₖ = 0.2). A force of 30 N is applied at 30° above horizontal. The acceleration is: (g = 10, sin30°=0.5, cos30°=√3/2≈0.866)
  1. A
    1.73 m/s²
  2. B
    1.27 m/s²
    Correct
  3. C
    3 m/s²
  4. D
    2 m/s²

Correct answer

1.27 m/s²

Explanation

N = mg − F sin30° = 100−15 = 85 N. Friction = 0.2×85 = 17 N. Net F_x = Fcos30°−friction = 30×0.866−17 = 25.98−17 = 8.98 N. a = 8.98/10 ≈ 0.9 m/s².

Review Laws of Motion and Friction notesStart timed chapter practice