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 step by step, we will break it down into two parts as specified in the question.

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

1. **Identify the Given Values**:
- Weight of the box (W) = 2000 N
- Distance (d) = 20 m
- Coefficient of friction (μ) = 0.2

2. **Draw the Free Body Diagram**:
- The box is being pulled at an angle θ with a force F.
- The vertical component of the force is \( F \sin \theta \).
- The horizontal component of the force is \( F \cos \theta \).

3. **Calculate the Normal Force (N)**:
- The normal force can be expressed as:
\[ N = W - F \sin \theta \]
- Since the box is moving slowly, the frictional force (f) can be calculated as:
\[ f = \mu N = \mu (W - F \sin \theta) \]

4. **Set Up the Equation for Horizontal Motion**:
- For the box to move, the horizontal component of the pulling force must equal the frictional force:
\[ F \cos \theta = \mu (W - F \sin \theta) \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ F \cos \theta + \mu F \sin \theta = \mu W \]
- Factoring out F:
\[ F (\cos \theta + \mu \sin \theta) = \mu W \]
- Thus, the force F can be expressed as:
\[ F = \frac{\mu W}{\cos \theta + \mu \sin \theta} \]

6. **Calculate the Work Done (W)**:
- The work done by the person pulling the box is given by:
\[ Work = F \cdot d \cdot \cos \theta \]
- Substituting for F:
\[ Work = \left(\frac{\mu W}{\cos \theta + \mu \sin \theta}\right) \cdot d \cdot \cos \theta \]
- Plugging in the values:
\[ Work = \left(\frac{0.2 \cdot 2000}{\cos \theta + 0.2 \sin \theta}\right) \cdot 20 \cdot \cos \theta \]
- Simplifying gives:
\[ Work = \frac{8000 \cos \theta}{\cos \theta + 0.2 \sin \theta} \]

### Part B: Finding the Work Done for Minimum Force

1. **Condition for Minimum Force**:
- To minimize the force, we set the derivative of \( \cos \theta + \mu \sin \theta \) to zero:
\[ \tan \theta = \mu \]
- Thus, \( \tan \theta = 0.2 \).

2. **Calculate Cosine and Sine Values**:
- Using \( \tan \theta = 0.2 \):
\[ \sin \theta = \frac{0.2}{\sqrt{1 + 0.2^2}} = \frac{0.2}{\sqrt{1.04}} \approx 0.196 \]
\[ \cos \theta = \frac{1}{\sqrt{1 + 0.2^2}} = \frac{1}{\sqrt{1.04}} \approx 0.98 \]

3. **Substituting Values into Work Equation**:
- Substitute \( \sin \theta \) and \( \cos \theta \) back into the work equation:
\[ Work = \frac{8000 \cdot 0.98}{0.98 + 0.2 \cdot 0.196} \]
- Calculate the denominator:
\[ 0.98 + 0.0392 \approx 1.0192 \]
- Thus, the work done becomes:
\[ Work \approx \frac{8000 \cdot 0.98}{1.0192} \approx 7680 \text{ J} \]

### Final Answers:
- **Part A**: Work done when pulling at angle θ is \( \frac{8000 \cos \theta}{\cos \theta + 0.2 \sin \theta} \).
- **Part B**: Work done for minimum force is approximately 7680 J.