A block of mass 10kg is placed on rough inclined plane of variable angle `theta` and friction coefficient `mu_(s)=mu_(k)=3//4`. When `theta` is `37^(@)` net reaction force applied by inclined is `vecN_(1)` and when `theta = 53^(@)` net reaction force applied by inclined is `vecN_(2)` , then `|vecN_(1)|-|vecN_(2)|` is :

Options

A. `0`
B. `25N`
C. `-25N`
D. `-37N`

Explanation

To solve the problem, we need to calculate the normal reaction forces \( \vec{N}_1 \) and \( \vec{N}_2 \) when the block is on the inclined plane at angles \( \theta = 37^\circ \) and \( \theta = 53^\circ \) respectively. We will then find the difference \( |\vec{N}_1| - |\vec{N}_2| \).

### Step 1: Calculate \( \vec{N}_1 \) at \( \theta = 37^\circ \)

1. **Identify the forces acting on the block**:
- Weight \( \vec{W} = mg \) acting downwards.
- Normal force \( \vec{N} \) acting perpendicular to the inclined plane.
- Frictional force \( \vec{F}_{friction} \) acting parallel to the inclined plane.

2. **Resolve the weight into components**:
- \( W = mg = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \)
- The component of weight parallel to the incline:
\[ W_{\parallel} = mg \sin \theta = 98 \sin(37^\circ) \]
- The component of weight perpendicular to the incline:
\[ W_{\perpendicular} = mg \cos \theta = 98 \cos(37^\circ) \]

3. **Calculate \( W_{\parallel} \) and \( W_{\perpendicular} \)**:
- Using \( \sin(37^\circ) \approx 0.6 \) and \( \cos(37^\circ) \approx 0.8 \):
\[ W_{\parallel} = 98 \times 0.6 = 58.8 \, \text{N} \]
\[ W_{\perpendicular} = 98 \times 0.8 = 78.4 \, \text{N} \]

4. **Calculate the normal force \( N_1 \)**:
- At \( \theta = 37^\circ \), the block is moving with constant speed, so the frictional force equals the component of weight parallel to the incline:
\[ F_{friction} = \mu_s N_1 \]
- Since \( \mu_s = \frac{3}{4} \):
\[ F_{friction} = \frac{3}{4} N_1 \]
- Setting the forces equal:
\[ W_{\parallel} = F_{friction} \implies 58.8 = \frac{3}{4} N_1 \]
- Solving for \( N_1 \):
\[ N_1 = \frac{58.8 \times 4}{3} = 78.4 \, \text{N} \]

### Step 2: Calculate \( \vec{N}_2 \) at \( \theta = 53^\circ \)

1. **Repeat the process for \( \theta = 53^\circ \)**:
- Using \( \sin(53^\circ) \approx 0.8 \) and \( \cos(53^\circ) \approx 0.6 \):
\[ W_{\parallel} = mg \sin(53^\circ) = 98 \times 0.8 = 78.4 \, \text{N} \]
\[ W_{\perpendicular} = mg \cos(53^\circ) = 98 \times 0.6 = 58.8 \, \text{N} \]

2. **Calculate the normal force \( N_2 \)**:
- Setting the forces equal:
\[ W_{\parallel} = F_{friction} \implies 78.4 = \frac{3}{4} N_2 \]
- Solving for \( N_2 \):
\[ N_2 = \frac{78.4 \times 4}{3} = 104.53 \, \text{N} \]

### Step 3: Calculate \( |\vec{N}_1| - |\vec{N}_2| \)

1. **Find the difference**:
\[ |\vec{N}_1| - |\vec{N}_2| = 78.4 - 104.53 = -26.13 \, \text{N} \]

### Final Answer:
\[ |\vec{N}_1| - |\vec{N}_2| = -26.13 \, \text{N} \]