JEE Main · 2005 · 08 May 2005 · Offline · Laws of Motion

A smooth block is released at rest on a

4 5^{\circ}

incline and then slides a distance d. The time taken to slide is

n

times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

Answer: (A)

Solution

On the smooth

4 5^{\circ}

incline the acceleration is

a_{1} = g sin 4 5^{\circ} = \frac{g}{2}

, and the block covers

d

in time

t_{1}

with

d = \frac{1}{2} a_{1} t_{1}^{2}

. On the rough incline friction cuts the acceleration to

a_{2} = g sin 4 5^{\circ} - μ_{k} g cos 4 5^{\circ} = \frac{g}{2} (1 - μ_{k})

, and the same

d

now takes

t_{2} = n t_{1}

, so

d = \frac{1}{2} a_{2} (n t_{1})^{2}

. As the distance is unchanged,

a_{1} t_{1}^{2} = a_{2} n^{2} t_{1}^{2}

, which gives

1 = (1 - μ_{k}) n^{2}

and hence

μ_{k} = 1 - \frac{1}{n ^{2}}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option