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Starting from rest, a body slides down a 45° inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

A

0.33

B

0.25

C

0.75

D

0.8

Step-by-Step Solution

  1. Smooth Incline (No Friction): The acceleration of a body sliding down a smooth incline is due to the component of gravity along the slope: a1=gsin⁡θa_1 = g \sin \theta. The time t1t_1 to cover distance ss is derived from s=12a1t12s = \frac{1}{2}a_1 t_1^2, giving t1=2sgsin⁡θt_1 = \sqrt{\frac{2s}{g \sin \theta}} [Source 56, 83].
  2. Rough Incline (With Friction): The acceleration is reduced by the opposing kinetic friction force (fk=μkmgcos⁡θf_k = \mu_k mg \cos \theta). The net acceleration is a2=gsin⁡θ−μkgcos⁡θ=g(sin⁡θ−μkcos⁡θ)a_2 = g \sin \theta - \mu_k g \cos \theta = g(\sin \theta - \mu_k \cos \theta) [Source 82]. The time t2t_2 is t2=2sa2t_2 = \sqrt{\frac{2s}{a_2}}.
  3. Relationship Given: The problem states t2=2t1t_2 = 2t_1. Squaring both sides: t22=4t12t_2^2 = 4t_1^2 2sa2=4(2sa1)  ⟹  a1=4a2\frac{2s}{a_2} = 4 \left( \frac{2s}{a_1} \right) \implies a_1 = 4a_2
  4. Substitution and Calculation: gsin⁡θ=4g(sin⁡θ−μkcos⁡θ)g \sin \theta = 4 g (\sin \theta - \mu_k \cos \theta) Cancel gg and rearrange: sin⁡θ=4sin⁡θ−4μkcos⁡θ\sin \theta = 4 \sin \theta - 4 \mu_k \cos \theta 4μkcos⁡θ=3sin⁡θ4 \mu_k \cos \theta = 3 \sin \theta μk=34tan⁡θ\mu_k = \frac{3}{4} \tan \theta Given θ=45∘\theta = 45^{\circ}, tan⁡45∘=1\tan 45^{\circ} = 1. μk=0.75\mu_k = 0.75
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