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Laws of Motion question

2015 · Q132
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Laws of Motion question

2015 · Q132

NEETPhysicsLaws of MotionMCQ+4 / −1
A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30o, the box starts to slip and slides 4.0 m down the plank in 4.0 s.

AIPMT 2015 Physics - Laws of Motion Question 25 English

The coefficients of static and kinetic friction between the box and the plank will be, respectively
  1. A
    0.5 and 0.6
  2. B
    0.4 and 0.3
  3. C
    0.6 and 0.6
  4. D
    0.6 and 0.5
View written solutionFree

Correct answer: D

Coefficient of static friction,
μs=tan⁡30∘=13=0.577≅0.6{\mu _s} = \tan 30^\circ = {1 \over {\sqrt 3 }} = 0.577 \cong 0.6μs​=tan30∘=3​1​=0.577≅0.6

S=ut+12at2S = ut + {1 \over 2}a{t^2}S=ut+21​at2

4=12a(4)2⇒a=12=0.54 = {1 \over 2}a{\left( 4 \right)^2} \Rightarrow a = {1 \over 2} = 0.54=21​a(4)2⇒a=21​=0.5
[∵\because∵ s = 4m and t = 4s given]

a = gsinθ\theta θ – μk(g)cos⁡θ{\mu _k}(g)\cos \theta μk​(g)cosθ

⇒μk=0.93=0.5 \Rightarrow {\mu _k} = {{0.9} \over {\sqrt 3 }} = 0.5⇒μk​=3​0.9​=0.5

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