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 ? .

Options

A. `(3)/(7cot theta)`
B. `(4)/(7cot theta)`
C. `(3)/(4cot theta)`
D. `(7)/(9cot theta)`

Answer

To solve the problem, we need to find the coefficient of friction between the ice and the incline. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have two scenarios:
1. An ice block sliding down a frictionless incline at a 45-degree angle.
2. The same ice block sliding down an incline with friction, taking twice as long to reach the bottom.

### Step 2: Determine the Acceleration on a Frictionless Incline
For the frictionless incline, the only force acting along the incline is the component of gravitational force. The acceleration \( a \) can be calculated as:
\[ a = g \sin \theta \]
For \( \theta = 45^\circ \):
\[ a = g \sin 45^\circ = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

### Step 3: Determine the Acceleration on the Frictional Incline
On the incline with friction, the forces acting are the gravitational force component down the incline and the frictional force opposing the motion. The net force \( F \) can be expressed as:
\[ F = mg \sin \theta - f \]
where \( f = \mu mg \cos \theta \) is the frictional force. Therefore, the net force becomes:
\[ F = mg \sin \theta - \mu mg \cos \theta \]
The acceleration \( a' \) can be expressed as:
\[ a' = \frac{mg \sin \theta - \mu mg \cos \theta}{m} = g \sin \theta - \mu g \cos \theta \]
Substituting \( \theta = 45^\circ \):
\[ a' = g \cdot \frac{1}{\sqrt{2}} - \mu g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}}(1 - \mu) \]

### Step 4: Relate the Times of Both Scenarios
Let \( t \) be the time taken to slide down the frictionless incline. The time taken to slide down the incline with friction is \( 2t \). The distance \( d \) traveled down the incline can be expressed using the equation of motion:
\[ d = \frac{1}{2} a t^2 \]
For the frictionless incline:
\[ d = \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t^2 \]
For the frictional incline:
\[ d = \frac{1}{2} a' (2t)^2 = 2 a' t^2 \]
Equating both expressions for distance:
\[ \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t^2 = 2 \left(\frac{g}{\sqrt{2}}(1 - \mu)\right) t^2 \]

### Step 5: Simplify the Equation
Cancelling \( t^2 \) from both sides (assuming \( t \neq 0 \)):
\[ \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) = 2 \left(\frac{g}{\sqrt{2}}(1 - \mu)\right) \]
Dividing both sides by \( \frac{g}{\sqrt{2}} \):
\[ \frac{1}{2} = 2(1 - \mu) \]
Solving for \( \mu \):
\[ \frac{1}{2} = 2 - 2\mu \implies 2\mu = 2 - \frac{1}{2} = \frac{3}{2} \implies \mu = \frac{3}{4} \]

### Final Answer
The coefficient of friction \( \mu \) between the ice and the incline is:
\[ \mu = \frac{3}{4} \]