A block of mass 50kg rests on a rough horizontal plane whose coefficient of static friction is `0.3`. What is the least force required to just move it if it is pulled (i) horizontally and (ii) at `45^(@)` to the horizontal.

Explanation

To solve the problem step by step, we need to determine the least force required to move a block of mass 50 kg resting on a rough horizontal plane with a coefficient of static friction of 0.3 in two scenarios: (i) when pulled horizontally and (ii) when pulled at an angle of 45 degrees to the horizontal.

### Step 1: Calculate the Normal Force (N)
The normal force (N) acting on the block can be calculated using the formula:
\[ N = mg \]
Where:
- \( m = 50 \, \text{kg} \) (mass of the block)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating:
\[ N = 50 \, \text{kg} \times 10 \, \text{m/s}^2 = 500 \, \text{N} \]

### Step 2: Calculate the Maximum Static Friction Force (f)
The maximum static friction force can be calculated using the formula:
\[ f = \mu_s N \]
Where:
- \( \mu_s = 0.3 \) (coefficient of static friction)

Calculating:
\[ f = 0.3 \times 500 \, \text{N} = 150 \, \text{N} \]

### Step 3: Case (i) - Force Required When Pulled Horizontally
When the block is pulled horizontally, the force required to overcome static friction is equal to the maximum static friction force calculated in Step 2.

Thus, the least force required to just move the block when pulled horizontally is:
\[ F = 150 \, \text{N} \]

### Step 4: Case (ii) - Force Required When Pulled at 45 Degrees
When the block is pulled at an angle of 45 degrees, we need to consider both the vertical and horizontal components of the applied force (F).

1. **Vertical Component**:
\[ F_y = F \sin(45^\circ) = \frac{F}{\sqrt{2}} \]

2. **Horizontal Component**:
\[ F_x = F \cos(45^\circ) = \frac{F}{\sqrt{2}} \]

Using the equilibrium of forces in the vertical direction:
\[ N + F_y = mg \]
Substituting for \( F_y \):
\[ N + \frac{F}{\sqrt{2}} = 500 \]

Rearranging gives:
\[ N = 500 - \frac{F}{\sqrt{2}} \]

Using the equilibrium of forces in the horizontal direction:
\[ F_x = f \]
Substituting for \( F_x \) and \( f \):
\[ \frac{F}{\sqrt{2}} = \mu_s N \]
Substituting \( N \):
\[ \frac{F}{\sqrt{2}} = 0.3 \left(500 - \frac{F}{\sqrt{2}}\right) \]

### Step 5: Solve for F
Expanding and rearranging the equation:
\[ \frac{F}{\sqrt{2}} = 150 - 0.3 \frac{F}{\sqrt{2}} \]
Combining like terms:
\[ \frac{F}{\sqrt{2}} + 0.3 \frac{F}{\sqrt{2}} = 150 \]
\[ \frac{1.3F}{\sqrt{2}} = 150 \]
Multiplying both sides by \( \sqrt{2} \):
\[ 1.3F = 150\sqrt{2} \]
Dividing by 1.3:
\[ F = \frac{150\sqrt{2}}{1.3} \]

Calculating \( F \):
\[ F \approx \frac{150 \times 1.414}{1.3} \approx 163.15 \, \text{N} \]

### Final Answers
1. The least force required to just move the block when pulled horizontally is **150 N**.
2. The least force required to just move the block when pulled at an angle of 45 degrees is approximately **163.15 N**.