JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Laws of Motion

A block starts moving up an inclined plane of inclination

3 0^{\circ}

with an initial velocity of

v_{0}

. It comes back to its initial position with velocity

\frac{v _{0}}{2}

. The value of the coefficient of kinetic friction between the block and the inclined plane is close to

\frac{I}{1000}

. The nearest integer to I is

Answer: 346

Solution

Going up, gravity and friction both oppose the motion, so

v_{0}^{2} = 2 g (sin 3 0^{\circ} + μ cos 3 0^{\circ}) S

, giving the distance

S = \frac{v _{0}^{2}}{g ( 1 + μ 3 )}

. Coming back down the same

S

from rest to

\frac{v _{0}}{2}

(\frac{v _{0}}{2})^{2} = 2 g (sin 3 0^{\circ} - μ cos 3 0^{\circ}) S

. Dividing the two relations clears

S

and

v_{0}

\frac{1}{4} = \frac{1 - μ 3}{1 + μ 3}

, so

1 + μ 3 = 4 (1 - μ 3)

and

5 μ 3 = 3

, giving

μ = \frac{3}{5 3} = \frac{3}{5} \approx 0.346

. Written as

\frac{I}{1000}

, the nearest integer is

I = 346

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Integer answer