JEE Main · 2014 · 06 Apr 2014 · Offline · Laws of Motion

A block of mass

m

is placed on a surface with a vertical cross section given by

y = x^{3} /6

. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is:

(A) $\frac{1}{3} m$
(B) $\frac{1}{2} m$
(C) $\frac{1}{6} m$ ✓ Correct
(D) $\frac{2}{3} m$

Answer: (C)

Solution

The block is on the point of sliding when friction is fully called upon, that is when

tan θ = μ = 0.5

, with

θ

the slope angle of the surface. The slope of the profile

y = x^{3} /6

\frac{d y}{d x} = \frac{x ^{2}}{2}

, so setting

\frac{x ^{2}}{2} = 0.5

gives

x^{2} = 1

, hence

x = 1

on the positive branch. The greatest height at which it can rest is then

y = \frac{x ^{3}}{6} = \frac{1}{6}

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option

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