A block of mass m is placed on a surface having vertical crossection given by y=x² / 4. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

Solution

Physics sir · 4 steps

On a curved surface the slope changes from point to point, so find where tan θ just reaches μ.

1. Condition for rest. A block on an incline of angle θ stays at rest as long as mg sin θ ≤ μ mg cos θ, that is tan θ ≤ μ. At the highest safe point, tan θ = μ.
2. Slope of the curve. At any point of the curve the incline angle is given by the slope of the tangent there. For y = x²/4 this slope grows as you move away from the bottom.

   tan θ = dy/dx = x/2 = μ = 0.5

3. Find the point. This gives x = 1 m. The height there is y = 1²/4 = 1/4 m.
4. So. The block can be placed up to a height of 1/4 m without slipping, which is option (d).

The idea to keep. For any curved surface y = f(x), the local incline is tan θ = dy/dx. Set it equal to μ to find the limiting x, then put that x back into f(x) for the height.

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