A block slides on ice with a velocity of `5m//s` comes to rest after moving through a distance of `13.5m` . Find the coefficient of friction ?

Explanation

To solve the problem of finding the coefficient of friction for a block sliding on ice, we can use the work-energy theorem. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
- Initial velocity of the block, \( v = 5 \, \text{m/s} \)
- Distance traveled before coming to rest, \( d = 13.5 \, \text{m} \)

### Step 2: Apply the Work-Energy Theorem
According to the work-energy theorem, the change in kinetic energy of the block is equal to the work done by the friction force.

The initial kinetic energy \( KE_i \) when the block is moving is given by:
\[ KE_i = \frac{1}{2} mv^2 \]
Since the block comes to rest, the final kinetic energy \( KE_f = 0 \).

### Step 3: Set Up the Equation
The work done by friction \( W_f \) can be expressed as:
\[ W_f = F_r \cdot d \]
where \( F_r \) is the friction force.

The friction force can be expressed as:
\[ F_r = \mu R \]
where \( \mu \) is the coefficient of friction and \( R \) is the normal force. On a horizontal surface, the normal force \( R \) is equal to the weight of the block, \( mg \).

Thus, we can rewrite the work done by friction as:
\[ W_f = \mu mg \cdot d \]

### Step 4: Equate Kinetic Energy to Work Done
From the work-energy theorem, we have:
\[ \frac{1}{2} mv^2 = \mu mg \cdot d \]
We can cancel the mass \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = \mu g \cdot d \]

### Step 5: Solve for the Coefficient of Friction \( \mu \)
Rearranging the equation gives:
\[ \mu = \frac{v^2}{2gd} \]

### Step 6: Substitute the Known Values
Substituting the values into the equation:
- \( v = 5 \, \text{m/s} \)
- \( g = 10 \, \text{m/s}^2 \) (approximate value of acceleration due to gravity)
- \( d = 13.5 \, \text{m} \)

We get:
\[ \mu = \frac{5^2}{2 \cdot 10 \cdot 13.5} = \frac{25}{270} \approx 0.0926 \]

### Final Answer
The coefficient of friction \( \mu \) is approximately \( 0.092 \).

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