A `10sqrt3` kg box has to move up an inclined slope of `60^(@)` to the horizontal at a uniform velocity of 5 `ms^(-1)` If the frictional force retarding the motion is 150N, the minimum force applied parallel to inclined plane to move up is (g=10 `ms^(-2)`)

Options

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

Answer

To solve the problem step by step, we will analyze the forces acting on the box as it moves up the inclined plane.

### Step 1: Identify the Given Data
- Mass of the box, \( m = 10\sqrt{3} \, \text{kg} \)
- Angle of incline, \( \theta = 60^\circ \)
- Uniform velocity, \( v = 5 \, \text{ms}^{-1} \)
- Frictional force, \( f_{\text{friction}} = 150 \, \text{N} \)
- Acceleration due to gravity, \( g = 10 \, \text{ms}^{-2} \)

### Step 2: Calculate the Weight of the Box
The weight \( W \) of the box can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the values:
\[ W = 10\sqrt{3} \cdot 10 = 100\sqrt{3} \, \text{N} \]

### Step 3: Resolve the Weight into Components
The weight can be resolved into two components:
1. Perpendicular to the incline: \( W_{\perpendicular} = W \cos \theta \)
2. Parallel to the incline: \( W_{\parallel} = W \sin \theta \)

Calculating these components:
- \( W_{\perpendicular} = 100\sqrt{3} \cdot \cos 60^\circ = 100\sqrt{3} \cdot \frac{1}{2} = 50\sqrt{3} \, \text{N} \)
- \( W_{\parallel} = 100\sqrt{3} \cdot \sin 60^\circ = 100\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 150 \, \text{N} \)

### Step 4: Apply Newton's Second Law for Uniform Velocity
Since the box is moving at a uniform velocity, the net force acting along the incline must be zero:
\[ F_{\text{applied}} - W_{\parallel} - f_{\text{friction}} = 0 \]
Rearranging gives:
\[ F_{\text{applied}} = W_{\parallel} + f_{\text{friction}} \]

### Step 5: Substitute the Values
Substituting the calculated values:
\[ F_{\text{applied}} = 150 \, \text{N} + 150 \, \text{N} = 300 \, \text{N} \]

### Step 6: Calculate the Minimum Force Required
To find the minimum force \( F \) applied parallel to the incline, we need to consider the angle:
\[ F \cos \theta = F_{\text{applied}} \]
Substituting \( \theta = 60^\circ \):
\[ F \cdot \frac{1}{2} = 300 \, \text{N} \]
Thus,
\[ F = 300 \cdot 2 = 600 \, \text{N} \]

### Final Answer
The minimum force applied parallel to the inclined plane to move the box up is:
\[ \boxed{600 \, \text{N}} \]