A block of mass 70 kg is kept on a rough horizontal surface and coefficient of static friction between block and surface is 0.4 . A man is trying to pull the block by applying a horizontal force .The net contact force exerted by the surface on the block is F , then:

Options

A. F must be 700 N
B. F must be 280 N
C. `700 N le F le 754 N`
D. F maybe greater than 754 N

Answer

To solve the problem step by step, we will analyze the forces acting on the block and calculate the required values.

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
- Weight (W) acting downwards: \( W = mg \)
- Normal force (N) acting upwards
- Applied force (F) acting horizontally
- Frictional force (f) acting opposite to the direction of the applied force

### Step 2: Calculate the weight of the block
Given the mass of the block \( m = 70 \, \text{kg} \) and acceleration due to gravity \( g = 10 \, \text{m/s}^2 \):
\[ W = mg = 70 \, \text{kg} \times 10 \, \text{m/s}^2 = 700 \, \text{N} \]

### Step 3: Determine the normal force
Since the block is on a horizontal surface and there are no vertical forces other than weight and normal force, the normal force (N) is equal to the weight of the block:
\[ N = W = 700 \, \text{N} \]

### Step 4: Calculate the maximum static frictional force
The maximum static frictional force can be calculated using the coefficient of static friction (\( \mu \)):
\[ f_{\text{max}} = \mu N = 0.4 \times 700 \, \text{N} = 280 \, \text{N} \]

### Step 5: Determine the range of the net contact force exerted by the surface on the block
The net contact force (F) exerted by the surface on the block will vary depending on the applied force (F). The range of F can be calculated as follows:
- The minimum contact force occurs when the applied force is zero (only the normal force acts):
\[ F_{\text{min}} = N = 700 \, \text{N} \]
- The maximum contact force occurs when the applied force is at its maximum static friction limit:
\[ F_{\text{max}} = \sqrt{N^2 + f_{\text{max}}^2} = \sqrt{700^2 + 280^2} \]
Calculating \( F_{\text{max}} \):
\[ F_{\text{max}} = \sqrt{490000 + 78400} = \sqrt{568400} \approx 754 \, \text{N} \]

### Step 6: Conclusion
Thus, the range of the net contact force exerted by the surface on the block is:
\[ 700 \, \text{N} < F < 754 \, \text{N} \]

### Final Answer
The range of F is from 700 N to 754 N.

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