A `30kg` box has to move up an inclined plane of slope `30^(@)` the horizontal with a unform velocity of `5 ms^(-1)`. If the frictional force retarding the motion is `150N`, the horizontal force required to move the box up is `(g =ms^(-2))` .

Options

A. `300xx(2)/sqrt3N`
B. `300xx(sqrt3)/(2)N`
C. `300N`
D. `150N`

Explanation

To solve the problem of finding the horizontal force required to move a 30 kg box up an inclined plane at an angle of 30 degrees with a uniform velocity of 5 m/s, we can follow these steps:

### Step 1: Identify the forces acting on the box
The forces acting on the box include:
- The weight of the box (mg), where m = 30 kg and g = 9.81 m/s².
- The component of the weight acting down the incline (mg sin θ).
- The normal force (N) acting perpendicular to the incline.
- The frictional force (F_friction) opposing the motion, which is given as 150 N.

### Step 2: Calculate the weight of the box
The weight (W) of the box can be calculated as:
\[ W = mg = 30 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 294.3 \, \text{N} \]

### Step 3: Calculate the components of the weight
The component of the weight acting down the incline (W_parallel) is given by:
\[ W_{\text{parallel}} = mg \sin \theta \]
Where θ = 30 degrees:
\[ W_{\text{parallel}} = 294.3 \, \text{N} \times \sin(30^\circ) = 294.3 \, \text{N} \times 0.5 = 147.15 \, \text{N} \]

The normal force (N) acting on the box is given by:
\[ N = mg \cos \theta \]
\[ N = 294.3 \, \text{N} \times \cos(30^\circ) = 294.3 \, \text{N} \times \frac{\sqrt{3}}{2} \approx 254.6 \, \text{N} \]

### Step 4: Calculate the frictional force
The frictional force is given as 150 N.

### Step 5: Set up the equation for the horizontal force
To move the box up the incline at a constant velocity, the total force acting up the incline must equal the sum of the forces acting down the incline (friction + component of weight):
\[ F_{\text{horizontal}} \cos \theta = W_{\text{parallel}} + F_{\text{friction}} \]
Substituting the known values:
\[ F_{\text{horizontal}} \cos(30^\circ) = 147.15 \, \text{N} + 150 \, \text{N} \]
\[ F_{\text{horizontal}} \cos(30^\circ) = 297.15 \, \text{N} \]

### Step 6: Solve for the horizontal force
Now, we can solve for \( F_{\text{horizontal}} \):
\[ F_{\text{horizontal}} = \frac{297.15 \, \text{N}}{\cos(30^\circ)} \]
Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ F_{\text{horizontal}} = \frac{297.15 \, \text{N}}{\frac{\sqrt{3}}{2}} = 297.15 \, \text{N} \times \frac{2}{\sqrt{3}} \approx 342.5 \, \text{N} \]

### Final Answer
Thus, the horizontal force required to move the box up the incline is approximately **342.5 N**.