A block of mass m is placed on a surface having vertical cross-section 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

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(3) 1/4 m. Idea: the surface gets steeper as you climb, so the block can rest anywhere the local slope is gentler than the angle of repose. The highest such point is where tan θ=μ exactly. The local slope is the derivative: tan θ=(dy)/(dx)=(2x)/4=x/2. Setting it equal to μ, x/2=0.5, so x=1 m. The height there is y=(x²)/4=1/4 m. Note the question asks for the height, not the horizontal distance: x=1 is the tempting wrong answer, and it is not even among the options. The 1/2 that is there is μ itself.

A block of mass m is placed on a surface having vertical cross-section 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

Step-by-step solution

Why the other options are wrong