A block of mass `2kg` is pushed down an inclined plane of angle `37^(@)` above horizontal with a force parallel to inclined plane. The coefficient of friction is 3/4. The block moves `10 m` in `2` seconds starting form rest. The magnitude of the force is

Options

A. `10 N`
B. `12 N`
C. `24 N`
D. `34 N`

Explanation

To solve the problem, we need to find the magnitude of the force applied to push the block down the inclined plane. We will follow these steps:

### Step 1: Identify the forces acting on the block
The forces acting on the block include:
1. The gravitational force acting downwards, \( F_g = mg \).
2. The normal force \( N \) acting perpendicular to the inclined plane.
3. The applied force \( F \) acting parallel to the inclined plane.
4. The frictional force \( f \) acting opposite to the direction of motion.

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

The gravitational force is:
\[ F_g = mg = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \]

### Step 3: Resolve the gravitational force into components
The component of the gravitational force acting down the incline is:
\[ F_{\text{gravity, parallel}} = F_g \sin(\theta) = 19.6 \, \text{N} \times \sin(37^\circ) \]
Using \( \sin(37^\circ) \approx 0.6 \):
\[ F_{\text{gravity, parallel}} \approx 19.6 \, \text{N} \times 0.6 \approx 11.76 \, \text{N} \]

The component of the gravitational force acting perpendicular to the incline is:
\[ F_{\text{gravity, perpendicular}} = F_g \cos(\theta) = 19.6 \, \text{N} \times \cos(37^\circ) \]
Using \( \cos(37^\circ) \approx 0.8 \):
\[ F_{\text{gravity, perpendicular}} \approx 19.6 \, \text{N} \times 0.8 \approx 15.68 \, \text{N} \]

### Step 4: Calculate the normal force
The normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[ N = F_{\text{gravity, perpendicular}} \approx 15.68 \, \text{N} \]

### Step 5: Calculate the frictional force
The frictional force \( f \) can be calculated using the coefficient of friction \( \mu = \frac{3}{4} = 0.75 \):
\[ f = \mu N = 0.75 \times 15.68 \, \text{N} \approx 11.76 \, \text{N} \]

### Step 6: Apply Newton's second law
Since the block moves down the incline, we can write the equation of motion along the incline:
\[ F - f - F_{\text{gravity, parallel}} = ma \]
Where:
- \( a \) is the acceleration of the block.
- The block moves \( 10 \, \text{m} \) in \( 2 \, \text{s} \), so we can find \( a \) using \( s = ut + \frac{1}{2} a t^2 \):
\[ 10 = 0 + \frac{1}{2} a (2^2) \implies 10 = 2a \implies a = 5 \, \text{m/s}^2 \]

### Step 7: Substitute values into the equation
Now substituting the known values into the equation:
\[ F - 11.76 \, \text{N} - 11.76 \, \text{N} = 2 \, \text{kg} \times 5 \, \text{m/s}^2 \]
\[ F - 23.52 \, \text{N} = 10 \, \text{N} \]
\[ F = 10 \, \text{N} + 23.52 \, \text{N} = 33.52 \, \text{N} \]

### Final Answer
The magnitude of the force \( F \) is approximately \( 33.52 \, \text{N} \).