A box weighing 2000 N is to be slowly slid through 20 m on a straigh track having friction coefficient 0.2 with the box. A find the work done by the person pulling the box wilth a chain at angle `theta` with the horizontal. B. find the work when the person has chosen a value of `theta` which ensures him the minimum magnitude of the force.

Explanation

To solve the problem, we need to break it down into two parts as specified in the question.

### Part A: Work Done by the Person Pulling the Box

1. **Identify the Forces Acting on the Box**:
- Weight of the box (W) = 2000 N (acting downward).
- Normal force (N) = acts perpendicular to the surface.
- Frictional force (F_friction) = μ * N, where μ = 0.2.
- Pulling force (F) at an angle θ with the horizontal.

2. **Calculate the Normal Force**:
- The vertical component of the pulling force is F * sin(θ).
- The normal force can be expressed as:
\[ N = W - F \cdot \sin(\theta) \]

3. **Calculate the Frictional Force**:
- The frictional force is given by:
\[ F_{friction} = \mu \cdot N = 0.2 \cdot (W - F \cdot \sin(\theta)) \]

4. **Determine the Horizontal Component of the Pulling Force**:
- The horizontal component of the pulling force is:
\[ F_{horizontal} = F \cdot \cos(\theta) \]

5. **Set Up the Equation for Equilibrium**:
- Since the box is moving slowly (constant velocity), the net force in the horizontal direction is zero:
\[ F_{horizontal} = F_{friction} \]
- Therefore:
\[ F \cdot \cos(\theta) = 0.2 \cdot (2000 - F \cdot \sin(\theta)) \]

6. **Solve for the Work Done**:
- Work done (W) is given by:
\[ W = F_{horizontal} \cdot d = F \cdot \cos(\theta) \cdot 20 \]
- Substitute \(F_{horizontal}\) from the equilibrium equation into the work done equation.

### Part B: Work Done at Minimum Force

1. **Find the Angle for Minimum Force**:
- The minimum force occurs when the frictional force is equal to the horizontal component of the pulling force. This happens when:
\[ \tan(\theta) = \mu \]
- Therefore:
\[ \theta = \tan^{-1}(0.2) \]

2. **Substitute θ into the Equations**:
- Substitute the value of θ back into the equations for normal force and frictional force to find the minimum force required.

3. **Calculate the Work Done**:
- Use the value of the minimum force in the work done equation:
\[ W_{min} = F_{horizontal} \cdot 20 \]

### Final Expressions

- For Part A, the work done can be expressed in terms of F and θ.
- For Part B, the work done will be calculated using the minimum force derived from the angle θ.