A force of 750 N is applied to a block of mass 102 kg to prevent it from sliding on a plane with an inclination angle `30^@` with the horizontal. If the coefficients of static friction and kinetic friction between the block and the plane are 0.4 and 0.3 respectively, then the frictional force acting on the block is

Options

A. 750 N
B. 500 N
C. 345 N
D. 250 N

Explanation

To solve the problem step by step, we will analyze the forces acting on the block and calculate the frictional force.

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
1. The gravitational force (weight) acting downward: \( W = mg \)
2. The applied force \( P = 750 \, \text{N} \) acting upward along the plane.
3. The frictional force \( f \) acting downward along the plane.
4. The normal force \( N \) acting perpendicular to the inclined plane.

### Step 2: Calculate the weight of the block
Given:
- Mass of the block, \( m = 102 \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

The weight of the block is calculated as:
\[ W = mg = 102 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 999.6 \, \text{N} \]

### Step 3: Resolve the weight into components
The weight can be resolved into two components:
1. Perpendicular to the inclined plane: \( W_{\perp} = mg \cos \theta \)
2. Parallel to the inclined plane: \( W_{\parallel} = mg \sin \theta \)

For \( \theta = 30^\circ \):
\[ W_{\perp} = mg \cos 30^\circ = 999.6 \, \text{N} \times \frac{\sqrt{3}}{2} \approx 866.0 \, \text{N} \]
\[ W_{\parallel} = mg \sin 30^\circ = 999.6 \, \text{N} \times \frac{1}{2} = 499.8 \, \text{N} \]

### Step 4: Calculate the net force acting on the block
The net force acting on the block along the incline can be expressed as:
\[ F_{\text{net}} = P - W_{\parallel} - f \]
Since the block is not sliding, we can set \( F_{\text{net}} = 0 \):
\[ 750 \, \text{N} - 499.8 \, \text{N} - f = 0 \]
This simplifies to:
\[ f = 750 \, \text{N} - 499.8 \, \text{N} = 250.2 \, \text{N} \]

### Step 5: Calculate the maximum static friction
The maximum static frictional force can be calculated using:
\[ f_{\text{max}} = \mu_s N \]
Where \( N = mg \cos \theta \) and \( \mu_s = 0.4 \):
\[ N = mg \cos 30^\circ = 999.6 \, \text{N} \times \frac{\sqrt{3}}{2} \approx 866.0 \, \text{N} \]
\[ f_{\text{max}} = 0.4 \times 866.0 \, \text{N} \approx 346.4 \, \text{N} \]

### Step 6: Determine the actual frictional force
Since the required frictional force \( f \) (250.2 N) is less than the maximum static friction (346.4 N), the actual frictional force acting on the block is equal to the required force to prevent sliding:
\[ f = 250.2 \, \text{N} \]

### Conclusion
The frictional force acting on the block is approximately **250 N**.