JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 3)
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 $${{{v_0}} \over 2}$$. The value of the coefficient of kinetic friction between
the block and the inclined plane is close to $${I \over {1000}}$$. The nearest integer to I is____.
Answer
346
Explanation
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a = g sin 30 + $$\mu $$ g cos 30
We know, v2 = u2 + 2as
$$ \Rightarrow $$ 0 = $$v_0^2$$ - 2ad
$$ \Rightarrow $$ $$v_0^2 = 2ad$$
$$d = {{v_0^2} \over {2a}}$$
Total work done,
$${W_f} = {k_f} - {k_i}$$
$$ \Rightarrow $$ $$ - 2\mu mg\,\cos 30{{v_0^2} \over {2a}} = {1 \over 2}m{{v_0^2} \over 4} - {1 \over 2}mv_0^2$$
$$ \Rightarrow $$ $${{ + \mu g\,\cos 30} \over a} = $$$${3 \over 8}$$
$$ \Rightarrow $$ $$8\mu g\,\cos 30 = 3g\,\sin 30 + 3\mu \,\cos 30$$
$$ \Rightarrow $$ $$5\mu g\,\cos 30 = 3g\,\sin 30$$
$$ \Rightarrow $$ $$\mu = {{3\tan 30} \over 5} = {{\sqrt 3 } \over 5}$$
$$ \Rightarrow $$ $${{\sqrt 3 } \over 5} = {I \over {1000}}$$
$$ \Rightarrow $$ $$I = 346$$
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