A block of mass 2 kg slides down an incline plane of inclination 30°. The coefficient of friction between block and plane is 0.5. The contact force between block and plank is :

Options

A. 20 N
B. `10sqrt3N`
C. `5sqrt7N`
D. `5sqrt15N`

Answer

To solve the problem of finding the contact force between a block of mass 2 kg sliding down an incline of 30° with a coefficient of friction of 0.5, we can follow these steps:

### Step 1: Identify the Forces Acting on the Block
The forces acting on the block include:
- The gravitational force (weight) \( W = mg \) acting vertically downward.
- The normal force \( N \) acting perpendicular to the incline.
- The frictional force \( f \) acting parallel to the incline, opposing the motion.

### Step 2: Calculate the Weight of the Block
Given:
- Mass \( m = 2 \) kg
- Acceleration due to gravity \( g = 10 \) m/s² (approximate value)

The weight \( W \) of the block is calculated as:
\[ W = mg = 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N} \]

### Step 3: Resolve the Weight into Components
The weight can be resolved into two components:
- Perpendicular to the incline: \( W_{\perp} = mg \cos(\theta) \)
- Parallel to the incline: \( W_{\parallel} = mg \sin(\theta) \)

For \( \theta = 30° \):
\[ W_{\perp} = 20 \cos(30°) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{N} \]
\[ W_{\parallel} = 20 \sin(30°) = 20 \times \frac{1}{2} = 10 \, \text{N} \]

### Step 4: Calculate the Normal Force
The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = W_{\perp} = 10\sqrt{3} \, \text{N} \]

### Step 5: Calculate the Frictional Force
The frictional force \( f \) can be calculated using the coefficient of friction \( \mu \):
\[ f = \mu N = 0.5 \times N = 0.5 \times 10\sqrt{3} = 5\sqrt{3} \, \text{N} \]

### Step 6: Calculate the Resultant Contact Force
The contact force \( F_c \) is the resultant of the normal force and the frictional force. Since these forces are perpendicular to each other, we can use the Pythagorean theorem:
\[ F_c = \sqrt{N^2 + f^2} \]
Substituting the values:
\[ F_c = \sqrt{(10\sqrt{3})^2 + (5\sqrt{3})^2} \]
\[ = \sqrt{100 \times 3 + 25 \times 3} = \sqrt{375} = 5\sqrt{15} \, \text{N} \]

### Final Answer
The contact force between the block and the incline is:
\[ F_c = 5\sqrt{15} \, \text{N} \]

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