A block starts moving up an inclined plane of inclination 30° with an initial velocity of v₀. It comes back to its initial position with velocity (v₀)/2. The value of the coefficient of kinetic friction between the block and the inclined plane is close to I/(1000). The nearest integer to I is ________.

Idea: the block covers the same distance up and down, so comparing the two accelerations through v²=2ad gives μ without ever needing the distance.

aᵤₚ=g(sin θ+μ cos θ), a_down=g(sin θ-μ cos θ).

Going up it loses all of v₀ over the distance d; coming back it gains (v₀)/2 over the same d:

v₀²=2aᵤₚd and (v₀²)/4=2a_downd.

Dividing, (a_down)/(aᵤₚ)=1/4:

4(sin θ-μ cos θ)=sin θ+μ cos θ

3 sin θ=5μ cos θ, so μ=3/5 tan θ.

At 30°, tan 30°=1/(√3):

μ=3/(5√3)=(√3)/5=0.3464.

Comparing with I/(1000) gives I=346.

Losing three-quarters of the kinetic energy to the round trip is a large loss, and μ=0.35 against tan 30°=0.58 is duly substantial.