Question

A piece of ice slides down a 45° incline in twice the time it takes to slide down a frictionless 45° incline. What is the coefficient of friction between the ice and the incline?

Answer 1

Here, θ = 45° ; $s_1$ = $s_2$ ; $u$ = 0

On the rough incline, $a_1$ = $g$ (sin θ - μ cos θ)

                                   $t_1$ = time taken

On the frictionless incline, $a_2$ = $g$ sin θ

            $t_2$ = time taken, and $t_1$ = 2 $t_2$

From       $s=ut+\dfrac{1}{2}at^2$

               $s_1=0+\dfrac{1}{2}\:g$ (sin θ - μ cos θ) $t_1^2$

and          $s_2 = 0+\dfrac{1}{2}\:g$ sin θ · $t_2^2$

As,           $s_1=s_2$

∴ $\dfrac{1}{2}\:g$ (sin θ - μ cos θ) $t_1^2$ = $\dfrac{1}{2}\:g$ sin θ · $t_2^2$

$\dfrac{\text{sin }\theta - \mu\text{ cos }\theta}{\text{sin }\theta} = \dfrac{t_2^2}{t_1^2}=\dfrac{t^2_2}{(2t_2)^2}=\dfrac{1}{4}$

     $1 - \mu\text{ cot }\theta = \dfrac{1}{4}$        or     $\mu\text{ cot }\theta=1-\dfrac{1}{4}=\dfrac{3}{4}$

               $\bold{\mu = \dfrac{3}{4\:cot\:\theta}}$