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Physics Question on laws of motion

A smooth block is released at rest on a 45∘^{\circ} incline and then slides a distance dd. The time taken to slide is nn times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

A

μk=1−1n2\mu_k = 1 - \frac{1}{n^2}

B

μk=1−1n2\mu_k = \sqrt{1- \frac{1}{n^2}}

C

μs=1−1n2\mu_s = 1 - \frac{1}{n^2}

D

μs=1−1n2\mu_s = \sqrt{1- \frac{1}{n^2}}

Answer

μk=1−1n2\mu_k = 1 - \frac{1}{n^2}

Explanation

Solution

When friction is absent
          a1=gsin⁡ θ\, \, \, \, \, \, \, \, \, \, a_1 = g \sin \, \theta
∴        s1=12a1t12            \therefore \, \, \, \, \, \, \, \, s_1 = \frac{1}{2 } a_1t_1^2 \, \, \, \, \, \, \, \, \, \, \, \, ...(i)
Whrn friction is present
           a2=gsin⁡ θ−μkgcos⁡θ\, \, \, \, \, \, \, \, \, \, \, a_2 = g \sin \, \theta - \mu_kg \cos \theta

∴          s2=12a2t22           \therefore \, \, \, \, \, \, \, \, \, \, s_2 = \frac{1}{2 a_2 t^2_2} \, \, \, \, \, \, \, \, \, \, \, ...(ii)
From Eqs. (i) and (ii)
or             12a1t12=12a2t12     ∵(t2=nt1) \, \, \, \, \, \, \, \, \, \, \, \, \frac{1}{2} a_1t^2_1 = \frac{1}{2} a_2t^2_1 \, \, \, \, \, \because (t_2 = nt_1)
or         a1=n2a2\, \, \, \, \, \, \, \, a_1 = n^2 a_2
        a2a1=gsin⁡θ−μkgcos⁡θgsin⁡θ=1n2\, \, \, \, \, \, \, \, \frac{a_2}{a_1} = \frac{g \sin \theta - \mu_k g \cos \theta}{g \sin \theta } = \frac{1}{n^2}
or         gsin⁡ 45∘−μkgcos⁡ 45∘g sin⁡ 45∘=1n2 \, \, \, \, \, \, \, \, \frac{g \sin \, 45^\circ - \mu_k g \cos \, 45^\circ}{g \, \sin \, 45^\circ } = \frac{1}{n^2}
or               1−μk=1n2\, \, \, \, \, \, \, \, \, \, \, \, \, \, 1 - \mu_k = \frac{1}{n^2} or μk=1−1n2\mu_k = 1 - \frac{1}{n^2}