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QJEEM20Ph5f69: JEE Main 2020 Physics - Class 11 JEEM20Ph5f69 Solution: A block starts moving up an inclined plane of inclination 30 o…

A block starts moving up an inclined plane of inclination 30o with an initial velocity of v0 . It comes back to its initial position with velocity v02{{{v_0}} \over 2} . The value of the coefficient of kinetic friction between the block and the inclined plane is close to I1000{I \over {1000}} . The nearest integer to I is____.

Step-by-Step Verified Solution

JEE Main 2020 (Online) 3rd September Evening Slot Physics - Work Power & Energy Question 89 English Explanation
a = g sin 30 + μ\mu g cos 30

We know, v2 = u2 + 2as

⇒\Rightarrow 0 = v02v_0^2 - 2ad

⇒\Rightarrow v02=2adv_0^2 = 2ad

d=v022ad = {{v_0^2} \over {2a}}

Total work done,

Wf=kf−ki{W_f} = {k_f} - {k_i}

⇒\Rightarrow −2μmg cos⁡30v022a=12mv024−12mv02- 2\mu mg\,\cos 30{{v_0^2} \over {2a}} = {1 \over 2}m{{v_0^2} \over 4} - {1 \over 2}mv_0^2

⇒\Rightarrow +μg cos⁡30a={{ + \mu g\,\cos 30} \over a} = 38{3 \over 8}

⇒\Rightarrow 8μg cos⁡30=3g sin⁡30+3μ cos⁡308\mu g\,\cos 30 = 3g\,\sin 30 + 3\mu \,\cos 30

⇒\Rightarrow 5μg cos⁡30=3g sin⁡305\mu g\,\cos 30 = 3g\,\sin 30

⇒\Rightarrow μ=3tan⁡305=35\mu = {{3\tan 30} \over 5} = {{\sqrt 3 } \over 5}

⇒\Rightarrow 35=I1000{{\sqrt 3 } \over 5} = {I \over {1000}}

⇒\Rightarrow I=346I = 346

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