A block starts moving up an inclined plane of inclination 30° with an initial velocity ofv_0. It comes back to its initial position with velocity v_0…
JEE Main 2020 — Physics Mechanics
2020integerhard
A block starts moving up an inclined plane of inclination 30∘ with an initial velocity ofv0. It comes back to its initial position with velocity 2v0. The value of the coefficient of kinetic friction between the block and the inclined plane is close to 10001, The nearest integer to I is :
Official previous-year question
Held on 3 Sept 2020 · Verified 6 Jul 2026.
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Solution
AtoB
a1=gsin30∘+μgcos30∘
=2g+2μg3;g=10ms2
v02−2a1(s)=0
s=a1v02 .....(i)
BtoA
a2=2g−2μ3g
(2V0)2=2a2(s)
s=4a2V02 .....(ii)
From equation (i) and (ii)
a1V02=4a2V02
⇒a1=4a2
⇒5+53μ=45−53μ
⇒5+53μ⇒253μ=15⇒μ=53=0.346=1000346
So, 10001=1000346
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