A body of mass 'm' is launched up on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of friction between the body and plane is √x/5 if the time of ascent is half of the time of descent. The value of x is __________.

Solution

Physics sir · 4 steps

Same distance up and down, so s = 1/2at² turns the time ratio into an a ratio.

1. Two accelerations. Going up, friction acts down the slope along with gravity, so the deceleration is a₁ = g(sin θ + μ cos θ). Coming down, friction acts up the slope, so a₂ = g(sin θ - μ cos θ).
2. Same distance. The ascent ends at rest, so read in reverse it is a start from rest: s = 1/2a₁ t₁². The descent also starts from rest: s = 1/2a₂ t₂². With t₂ = 2t₁, this gives a₁ = 4a₂.

   sin θ + μ cos θ = 4(sin θ - μ cos θ) ⇒ μ = 3/5 tan θ

3. Put in the angle. With θ = 30°, μ = 3/5 × 1/√3 = √3/5. Comparing with √x/5 gives √x = √3.
4. So. x = 3.

The idea to keep. Friction adds to gravity going up and opposes it coming down. Over equal distances from rest, t² ∝ 1/a, so a time ratio of 2 means an acceleration ratio of 4.

Answer from the NTA final answer key. Solution by PrepWiser.