Final answer:
The maximum height at which a block can be placed without slipping on a surface with a vertical cross-section y=x^3/6, with a coefficient of static friction of 0.5, is found to be 1/3 m.
"The correct option is approximately option A"
Step-by-step explanation:
The question asks about the maximum height above the ground at which a block can be placed on a surface with a given equation without slipping, given a coefficient of static friction of 0.5.
To solve for the maximum height, we must first understand the relationship between the slope of the surface and the static friction. The equation y = x^3/6 represents the vertical cross-section of the surface.
The slope of the surface at any point is given by the derivative dy/dx, which equals x^2/2.
The maximum height will occur when the static frictional force is equal to the component of gravitational force parallel to the slope, which can be calculated using the formula fs_max = μ_s * N, with μ_s being the coefficient of static friction and N the normal force. Since the block is not moving, N = mg, and the maximum static frictional force is 0.5mg. The gravitational force component parallel to the slope at the point where slipping begins is mg * sin(θ), where θ can be obtained from the slope x^2/2 at that point. Equating these forces and simplifying gives us the maximum height as 1/3 m, so the correct answer is A. 1/3 m.