Problem:
A smooth block is released at rest on a 45-degree incline and then slides a distance d. The time taken to slide n times as much on a rough incline than on a smooth incline. The coefficient of friction is?

Solution:
To solve this problem, we need to analyze the motion of the block on both the smooth and rough inclines. Let's break it down step by step:

1. Motion on a smooth incline:
When the block is released on a smooth incline, there is no friction acting on it. The only force acting on the block is its weight, which can be resolved into two components: one parallel to the incline (mg*sinθ) and one perpendicular to the incline (mg*cosθ). Here, θ is the angle of the incline.

The parallel component of weight (mg*sinθ) is responsible for the acceleration of the block down the incline. The equation for acceleration on an incline is given by a = g*sinθ, where g is the acceleration due to gravity.

Using the equation of motion, s = ut + (1/2)at^2, where u is the initial velocity, t is the time taken, and s is the distance traveled, we can write:

d = 0 + (1/2)(g*sinθ)t^2
d = (1/2)gt^2*sinθ
2d/g*sinθ = t^2
t = √(2d/g*sinθ)

2. Motion on a rough incline:
When the block is released on a rough incline, there is an additional force acting on it due to friction. The friction force opposes the motion of the block and can be calculated using the equation f = μN, where μ is the coefficient of friction and N is the normal force.

The normal force (N) can be calculated as N = mg*cosθ, where m is the mass of the block.

The net force acting on the block down the incline is given by F_net = mg*sinθ - μN. Using Newton's second law, F_net = ma, we can write:

mg*sinθ - μN = ma
mg*sinθ - μmg*cosθ = ma
g*sinθ - μg*cosθ = a

Now, using the equation of motion, s = ut + (1/2)at^2, we can write:

n*d = 0 + (1/2)(g*sinθ - μg*cosθ)t^2
2n*d = (g*sinθ - μg*cosθ)t^2
t = √(2n*d/(g*sinθ - μg*cosθ))

3. Comparing the times:
The problem states that the time taken to slide n times as much on a rough incline compared to a smooth incline is the same as the time taken to slide on a smooth incline. Mathematically, this can be written as:

√(2d/g*sinθ) = √(2n*d/(g*sinθ - μg*cosθ))

Squaring both sides of the equation and simplifying, we get:

2d/(g*sinθ) = 2n*d/(g*sinθ - μg*cosθ)
(g*sinθ - μg*cos

1-1/n^2