A `3 kg` block of wood is on a level surface where `mu_(s) = 0.25` and `mu_(s) = 0.2 `A force of `7 N` is being applied horizontal to the block Mark the correct statement(s) regarding this situation.

Options

A. If the block is initially at rest , it will remain at rest and friction force will be about ?`7 N`
B. If the block is initially moving than it will continue its motion forever if the force applies is in the direction of the motion of the block
C. If the block is initially moving and the direction of applied force is mass as that of motion of block , than block moves with an acceleration of `1//3 ms^(-2)` along its initial direction of motion
D. If the block is initially moving and direction of applied force to that of initial motion of the blockthan the block decelerates comes to a stop and starts moving in the opposite direction.

Explanation

To solve the problem, we need to analyze the situation involving a 3 kg block of wood on a level surface with given coefficients of static and kinetic friction. We will determine the correct statements regarding the block's motion under the influence of an applied force of 7 N.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The weight of the block (W) can be calculated using the formula:
\[ W = m \cdot g \]
where \( m = 3 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
\[ W = 3 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 29.4 \, \text{N} \]
- The normal force (N) acting on the block is equal to its weight since the surface is level:
\[ N = W = 29.4 \, \text{N} \]

2. **Calculate Maximum Static Friction (F_s_max):**
- The maximum static friction can be calculated using the coefficient of static friction (\( \mu_s = 0.25 \)):
\[ F_{s_{\text{max}}} = \mu_s \cdot N = 0.25 \times 29.4 \, \text{N} = 7.35 \, \text{N} \]

3. **Calculate Kinetic Friction (F_k):**
- The kinetic friction can be calculated using the coefficient of kinetic friction (\( \mu_k = 0.2 \)):
\[ F_k = \mu_k \cdot N = 0.2 \times 29.4 \, \text{N} = 5.88 \, \text{N} \]

4. **Analyze the Applied Force:**
- An applied force of 7 N is acting on the block. Since \( 7 \, \text{N} < 7.35 \, \text{N} \), the block will remain at rest because the applied force does not exceed the maximum static friction.

5. **Evaluate the Statements:**
- **Statement A:** If the block is initially at rest, it will remain at rest. This statement is **true** because the applied force (7 N) is less than the maximum static friction (7.35 N).
- **Statement B:** If the block is initially moving, it will continue its motion forever if the force applied is in the direction of motion. This statement is **true** because once in motion, the applied force (7 N) exceeds the kinetic friction (5.88 N), allowing continuous motion.
- **Statement C:** If the block is initially moving in the direction of the applied force, it will move with an acceleration of \( \frac{1}{3} \, \text{m/s}^2 \). This statement is **true** because the net force when moving is \( 7 \, \text{N} - 5.88 \, \text{N} = 1.12 \, \text{N} \), leading to an acceleration of:
\[ a = \frac{F_{\text{net}}}{m} = \frac{1.12 \, \text{N}}{3 \, \text{kg}} \approx 0.373 \, \text{m/s}^2 \]
- **Statement D:** If the block is initially moving in the direction of the applied force, it will decelerate and stop, then move in the opposite direction. This statement is **false** because the block will not reverse direction unless the applied force is less than the opposing friction force.

### Conclusion:
The correct statements are A, B, and C.