A heavy body of mass `25kg` is to be dragged along a horizontal plane (`mu` = `(1)/(sqrt(3))`. The least force required is `(1 kgf = 9.8 N)`

Explanation

To solve the problem of dragging a heavy body of mass 25 kg along a horizontal plane with a coefficient of friction (μ) of \( \frac{1}{\sqrt{3}} \), we will follow these steps:

### Step 1: Identify the forces acting on the body
The forces acting on the body include:
- The gravitational force (weight) acting downwards, \( W = mg \)
- The normal force (N) acting upwards
- The frictional force (F_f) acting opposite to the direction of motion
- The applied force (F) required to drag the body

### Step 2: Calculate the weight of the body
The weight \( W \) of the body can be calculated using the formula:
\[ W = mg \]
where:
- \( m = 25 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)

Calculating \( W \):
\[ W = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 245 \, \text{N} \]

### Step 3: Determine the normal force
On a horizontal plane, the normal force \( N \) is equal to the weight of the body:
\[ N = W = 245 \, \text{N} \]

### Step 4: Calculate the frictional force
The frictional force \( F_f \) can be calculated using the formula:
\[ F_f = \mu N \]
where:
- \( \mu = \frac{1}{\sqrt{3}} \)

Calculating \( F_f \):
\[ F_f = \frac{1}{\sqrt{3}} \times 245 \, \text{N} = \frac{245}{\sqrt{3}} \, \text{N} \]

### Step 5: Calculate the least force required to move the body
The least force \( F \) required to overcome friction and move the body is equal to the frictional force:
\[ F = F_f = \frac{245}{\sqrt{3}} \, \text{N} \]

### Step 6: Convert the force to kgf
To convert the force from Newtons to kgf, we use the conversion \( 1 \, \text{kgf} = 9.8 \, \text{N} \):
\[ F_{\text{kgf}} = \frac{F}{9.8} = \frac{245/\sqrt{3}}{9.8} \, \text{kgf} \]

### Final Answer
The least force required to drag the body is:
\[ F = \frac{245}{\sqrt{3}} \, \text{N} \quad \text{or} \quad \approx 141.45 \, \text{N} \]