A block starts moving up a fixed inclined plane of inclination `60^(@)C` with a velocityof `20 ms^(-1)` and stops after 2s. The approximate value of the coefficient of friction is `[g = 10 ms^(-2)`]

Options

A. 3
B. 3.3
C. 0.27
D. 0.33

Answer

To solve the problem step by step, we will analyze the forces acting on the block moving up the inclined plane and use kinematics to find the coefficient of friction.

### Step 1: Identify the forces acting on the block
The forces acting on the block moving up the inclined plane are:
- The gravitational force (weight) acting downward, \( mg \).
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( F_k \) acting opposite to the direction of motion (down the incline).

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

Given that \( \theta = 60^\circ \):
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
- \( \cos 60^\circ = \frac{1}{2} \)

Thus, the components are:
- \( mg \sin 60^\circ = mg \cdot \frac{\sqrt{3}}{2} \)
- \( mg \cos 60^\circ = mg \cdot \frac{1}{2} \)

### Step 3: Write the equation of motion
Since the block is moving up the incline and then stops, the net force acting on it can be expressed as:
\[ F_{\text{net}} = -mg \sin 60^\circ - F_k \]
Where \( F_k = \mu N \) and \( N = mg \cos 60^\circ \).

Substituting for \( N \):
\[ F_k = \mu (mg \cos 60^\circ) = \mu \left( mg \cdot \frac{1}{2} \right) \]

### Step 4: Set up the equation
The net force can also be expressed in terms of mass and acceleration:
\[ F_{\text{net}} = ma \]
Thus, we have:
\[ -mg \sin 60^\circ - \mu \left( mg \cdot \frac{1}{2} \right) = ma \]

Since \( a \) is negative (retardation), we can express it as:
\[ a = -g \left( \frac{1}{2} \sqrt{3} + \frac{\mu}{2} \right) \]

### Step 5: Use kinematics to find acceleration
Using the kinematic equation:
\[ v = u + at \]
Where:
- Final velocity \( v = 0 \) (the block stops)
- Initial velocity \( u = 20 \, \text{m/s} \)
- Time \( t = 2 \, \text{s} \)

Substituting the values:
\[ 0 = 20 + \left(-g \left( \frac{1}{2} \sqrt{3} + \frac{\mu}{2} \right)\right) \cdot 2 \]

### Step 6: Solve for \( \mu \)
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ 0 = 20 - 10 \left( \frac{1}{2} \sqrt{3} + \frac{\mu}{2} \right) \cdot 2 \]
\[ 0 = 20 - 10 \left( \sqrt{3} + \mu \right) \]
\[ 10 \left( \sqrt{3} + \mu \right) = 20 \]
\[ \sqrt{3} + \mu = 2 \]
\[ \mu = 2 - \sqrt{3} \]

### Step 7: Calculate the approximate value of \( \mu \)
Using the approximate value \( \sqrt{3} \approx 1.732 \):
\[ \mu \approx 2 - 1.732 = 0.268 \]

Thus, the approximate value of the coefficient of friction \( \mu \) is approximately \( 0.268 \).