a block of weight 100 N is slowly slid up on a smooth incline of inclination `37^0` by as person. Calculate the work done by the person in moving the block through a distance of 2.0 m, if the drivin force is a. parallel to the inclne and b. in the horizontla direction.

Explanation

To solve the problem, we need to calculate the work done by a person in moving a block of weight 100 N up a smooth incline of inclination \(37^\circ\) through a distance of 2.0 m. We will consider two cases: (a) when the driving force is parallel to the incline, and (b) when the driving force is in the horizontal direction.

### Step-by-Step Solution

**Case (a): Driving Force Parallel to the Incline**

1. **Identify the Forces Acting on the Block:**
- The weight of the block \(W = 100 \, \text{N}\).
- The angle of inclination \(\theta = 37^\circ\).
- The component of the weight acting down the incline is \(W \sin(\theta)\).
- The component of the weight acting perpendicular to the incline is \(W \cos(\theta)\).

2. **Calculate the Component of Weight Down the Incline:**
\[ W \sin(37^\circ) = 100 \sin(37^\circ) \]
Using \(\sin(37^\circ) \approx 0.6\):
\[ W \sin(37^\circ) = 100 \times 0.6 = 60 \, \text{N} \]

3. **Determine the Force Required to Move the Block Up:**
- Since the block is moved slowly up the incline, the person must apply a force \(F\) that is slightly greater than \(60 \, \text{N}\) to overcome the component of weight acting down the incline. We can assume \(F \approx 60 \, \text{N}\) for our calculations.

4. **Calculate the Work Done:**
- The work done \(W_d\) is given by:
\[ W_d = F \times d \times \cos(0^\circ) \]
where \(d = 2 \, \text{m}\) is the distance moved and \(\cos(0^\circ) = 1\).
\[ W_d = 60 \, \text{N} \times 2 \, \text{m} \times 1 = 120 \, \text{J} \]

**Case (b): Driving Force in the Horizontal Direction**

1. **Identify the Forces Acting on the Block:**
- The weight of the block remains \(W = 100 \, \text{N}\).
- The horizontal force \(F_h\) is applied.

2. **Calculate the Component of the Horizontal Force:**
- The horizontal force has a component that acts along the incline:
\[ F_h \cos(37^\circ) = 100 \sin(37^\circ) \]
Using \(\cos(37^\circ) \approx 0.8\):
\[ F_h \cos(37^\circ) = 100 \times 0.6 = 60 \, \text{N} \]
Therefore,
\[ F_h = \frac{60 \, \text{N}}{\cos(37^\circ)} = \frac{60 \, \text{N}}{0.8} = 75 \, \text{N} \]

3. **Calculate the Work Done:**
- The work done when applying the force horizontally is given by:
\[ W_d = F_h \times d \times \cos(53^\circ) \]
where \(d = 2 \, \text{m}\) and \(\cos(53^\circ) \approx 0.6\).
\[ W_d = 75 \, \text{N} \times 2 \, \text{m} \times 0.6 = 90 \, \text{J} \]

### Final Answers
- **Work done when the force is parallel to the incline:** \(120 \, \text{J}\)
- **Work done when the force is horizontal:** \(90 \, \text{J}\)