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

The correct answer is 0.75.

We are told that a body, starting from rest, slides down the same distance on a 45° inclined plane, but it takes twice as long when friction is present compared to when there is no friction. Let's call:

t₁ = time to slide down without friction
t₂ = time to slide down with friction = 2t₁

Step 1: Find acceleration without friction

When there is no friction, the only force along the incline is the component of gravity:

a_1=gsin\mathrm{45°}=\frac{g}{\sqrt{2}}

Step 2: Find acceleration with friction

When friction is present, the net acceleration down the incline is reduced by the friction force (which acts up the incline):

a_2=gsin\mathrm{45°}-\mu gcos\mathrm{45°}=g(sin\mathrm{45°}-\mu cos\mathrm{45°})

Since sin 45° = cos 45° = 1/√2, this simplifies to:

a_2=\frac{g(1-\mu )}{\sqrt{2}}

Step 3: Apply the kinematic equation for the same distance

Since the body starts from rest and travels the same distance s in both cases, using s=\frac{1}{2}a{t}^2:

\frac{1}{2}{a}_1{t}_1^2=\frac{1}{2}{a}_2{t}_2^2

Cancelling the ½:

a_1{t}_1^2=a_2{t}_2^2

Step 4: Substitute t₂ = 2t₁

a_1{t}_1^2=a_2\left(2t_1\right){}^2=4a_2{t}_1^2

Dividing both sides by t₁²:

a_1=4a_2

Step 5: Solve for the coefficient of friction μ

Substituting the expressions for a₁ and a₂:

\frac{g}{\sqrt{2}}=4\times \frac{g(1-\mu )}{\sqrt{2}}

Multiplying both sides by √2 and dividing by g:

1=4(1-\mu )

1=4-4\mu

4\mu =4-1=3

\mu =\frac{3}{4}=0.75

Therefore, the coefficient of friction between the body and the inclined plane is 0.75. The key insight is that since the same distance is covered in twice the time (with friction vs. without), the acceleration with friction must be exactly one-fourth of the acceleration without friction (since distance ∝ t² for motion from rest, and (2t)² = 4t²).