Question

Chapter :- Laws of motion

A block of mass is placed on a surface with a vertical cross -section given y = x³/6 . If the cofficient of friction is 0.5 , the maximum height above the ground at which the can be placed without slipping is ?
[JEE main 2014]

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Answer 1

Explanation :- A block of mass m is placed on a surface with a vertical cross section , then

tan = dy/dx = d(x³/6)/dx = x²/2'

• At equilibrium , we get
• μ = tanθ
• 0.5 = x²/2
• x² = 1 => x = +-1

Now , putting the value of x in y = x³/6 , we get

when x = 1

• ∴ y = 1/6

when x = -1

• y = -1/6

so , the maximum height above the ground at which the block can be placed without slipping is 1/6m Answer

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