In the post dated 9^{th} August 2007, two questions on friction were given without solution. These questions with solution are given below as promised in the post:

**(1) A block of mass ‘M’ equal to 1 kg is resting on a frictionless horizontal surface. Another block of mass ‘m’ equal to 0.2 kg is resting on the top surface of the 1 kg block. The coefficient of friction between the two blocks is 0.2. The motion of the 1 kg kg block is controlled by a light spring of force constant 24 Nm**^{–1}** arranged as shown**** in the figure. What is the maximum amplitude of oscillation of the blocks (together) so that the upper block does not slip relative to the lower block? (g = 10 ms**^{–2}**)**

**(a) 2 cm (b) 5 cm (c) 10 cm (d) 20 cm (e)50cm **

The figure is not given here. You will find it with the post dated 9^{th} August 2007 by moving a little down on this page.

If ‘x’ is the *maximum* possible amplitude, the maximum restoring force exeretred by the spring is Kx and the maximum acceleration of the two blocks (while moving together) is Kx/(M+m).

The force experienced by the upper block while moving with the above acceleration is mKx/(M+m). The upper block will just start slipping when this force exceeds the frictional force μmg where ‘μ’ is the coefficient of friction so that in the limiting case, we have

mKx/(M+m) = μmg.

Therefore, x = μg(M+m)/K = 0.2×10(1+0.2)/24 = 0.1 m = 10 cm.

(2) A metallic cube takes ‘t’ seconds to slide down a smooth inclined plane of angle 45º. When the inclined plane is made rough, the block takes nt

**(a) 1/n (b) 1 ****–**** (1/n) (c) 1 ****–**** (1/n ^{2}) (d) (1/n^{2}) (e) 1 **

**+**

**(1/n**

^{2})In the case of the smooth inclined plane, the acceleration of the block down the plane is gsinθ where ‘θ’ is the angle of the plane (equal to 45º). If ‘s’ is the length of the plane, we have

s = 0 + ½ (gsinθ)t^{2}

[We have used the equation of linear motion, s = ut + ½ at^{2}]

In the case of the rough inclined plane, since the net force acting on the block is (mgsinθ – μmgcosθ), the acceleration of the block down the plane is g(sinθ – μcosθ).

Therefore, we have

s = 0 + ½ g(sinθ – μcosθ) (nt)^{2}

From the above equations, we obtain

sinθ(n^{2} – 1) = μn^{2}cosθ

Therefore, μ = [(n^{2} – 1) /n^{2}] tanθ.

Since θ = 45º, tanθ =1 and hence μ = (n^{2} – 1)/n^{2} = **1 ****–**** (1/n ^{2}) **

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