NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
The figure below shows a block of mass 15 kg , kept on a rough inclined plane of angle 30 ° and coefficient of static friction equal to 0 . 5 . It is being acted upon by two forces. What should be the minimum value of P (in N ) so that the block doesn't slip downwards? [ Take g = 10 ms - 2 and 3 = 1 . 7 ]
Correct answer
16.25
Step-by-step solution
F + mg sin θ = P + μ mg cos θ ∴ 5 + 15 × 10 × 1 2 = P = 0.5 × 15 × 10 × 3 2 ∴ 80 = P + 75 3 2 ∴ P = 80 − 75 3 2 ∴ P = 160 − 75 3 2 ∴ P = 16.25 N