A block is released from rest from point on a rough inclined place of inclination `37^(@).` The coefficient of friction is `0.5.`

Options

A. The time taken to slide down 9 m on the plane is 3 s.
B. The velocity of block after moving 4 m is m/s.
C. The block travels equal distances in equal intervals of time.
D. The velocity of block increases linearly.

Answer

To solve the problem step by step, we will analyze the forces acting on the block on the inclined plane and then calculate the acceleration, time taken to slide down a certain distance, and the velocity after moving a specific distance.

### Step 1: Identify the Forces Acting on the Block
1. **Weight of the Block (W)**: The weight acts vertically downward and is given by \( W = mg \).
2. **Components of Weight**:
- **Parallel to the incline**: \( W_{\parallel} = mg \sin \theta \)
- **Perpendicular to the incline**: \( W_{\perpendicular} = mg \cos \theta \)
3. **Normal Force (N)**: The normal force acts perpendicular to the incline and is equal to the perpendicular component of the weight: \( N = mg \cos \theta \).
4. **Frictional Force (F_f)**: The frictional force opposes the motion and is given by \( F_f = \mu N = \mu mg \cos \theta \).

### Step 2: Write the Equation of Motion
Since the block is sliding down the incline, we can write the equation of motion along the incline:
\[ ma = mg \sin \theta - F_f \]
Substituting the expression for frictional force:
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
Dividing through by \( m \):
\[ a = g \sin \theta - \mu g \cos \theta \]

### Step 3: Substitute Known Values
Given:
- \( \theta = 37^\circ \)
- \( \mu = 0.5 \)
- \( g \approx 10 \, \text{m/s}^2 \)

Calculate \( \sin 37^\circ \) and \( \cos 37^\circ \):
- \( \sin 37^\circ = \frac{3}{5} \)
- \( \cos 37^\circ = \frac{4}{5} \)

Now substitute these values into the equation for acceleration:
\[ a = 10 \left( \frac{3}{5} \right) - 0.5 \times 10 \left( \frac{4}{5} \right) \]
\[ a = 6 - 4 = 2 \, \text{m/s}^2 \]

### Step 4: Calculate Time to Slide Down 9 Meters
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the block is released from rest, \( u = 0 \):
\[ 9 = 0 + \frac{1}{2} \times 2 \times t^2 \]
\[ 9 = t^2 \]
\[ t = 3 \, \text{s} \]

### Step 5: Calculate Velocity After Moving 4 Meters
Using the equation:
\[ v^2 = u^2 + 2as \]
Again, since \( u = 0 \):
\[ v^2 = 0 + 2 \times 2 \times 4 \]
\[ v^2 = 16 \implies v = 4 \, \text{m/s} \]

### Step 6: Analyze Distance Travelled in Equal Time Intervals
1. **Distance in the first second (s1)**:
\[ s_1 = \frac{1}{2} a t^2 = \frac{1}{2} \times 2 \times 1^2 = 1 \, \text{m} \]
2. **Distance in the first two seconds (s2)**:
\[ s_2 = \frac{1}{2} a t^2 = \frac{1}{2} \times 2 \times 2^2 = 4 \, \text{m} \]
3. **Distance travelled in the second second (s2 - s1)**:
\[ s_2 - s_1 = 4 - 1 = 3 \, \text{m} \]
The distances are not equal, therefore the statement is incorrect.

### Step 7: Determine if Velocity Increases Linearly
Since the acceleration is constant (\( a = 2 \, \text{m/s}^2 \)), the velocity increases linearly with time.

### Summary of Results
- Acceleration of the block: \( 2 \, \text{m/s}^2 \)
- Time taken to slide down 9 meters: \( 3 \, \text{s} \)
- Velocity after moving 4 meters: \( 4 \, \text{m/s} \)
- The block does not travel equal distances in equal intervals of time.
- The velocity of the block increases linearly.