A block of weight 100 N is slowly slide 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 driving force is a. parallel to the incline and b. in the horizontal direction.

Explanation

To solve the problem, we need to calculate the work done by a person in two scenarios: when the driving force is parallel to the incline and when it is in the horizontal direction.

### Given Data:
- Weight of the block (W) = 100 N
- Distance moved (d) = 2.0 m
- Angle of inclination (θ) = 37°

### Part (a): Work Done When the Force is Parallel to the Incline

1. **Calculate the component of weight acting down the incline**:
The force due to gravity acting down the incline can be calculated using the formula:
\[ F_{\text{gravity}} = W \sin(\theta) \]
Substituting the values:
\[ F_{\text{gravity}} = 100 \sin(37^\circ) \]
Using \(\sin(37^\circ) \approx \frac{3}{5} = 0.6\):
\[ F_{\text{gravity}} = 100 \times 0.6 = 60 \text{ N} \]

2. **Calculate the work done by the person**:
Since the person is moving the block up the incline slowly, the force exerted by the person must equal the component of the weight acting down the incline. Therefore, the force exerted by the person (F) is also 60 N.
The work done (W) is given by:
\[ W = F \times d \]
Substituting the values:
\[ W = 60 \times 2 = 120 \text{ J} \]

### Part (b): Work Done When the Force is in the Horizontal Direction

1. **Identify the components of the forces**:
When the force is applied horizontally, we need to consider the incline's angle. The horizontal force will have two components: one along the incline and one perpendicular to it. The component of the weight acting down the incline remains the same at 60 N.

2. **Calculate the effective force along the incline**:
The horizontal force (F_horizontal) can be resolved into components. The component of the horizontal force that acts parallel to the incline can be calculated using:
\[ F_{\text{parallel}} = F_{\text{horizontal}} \cos(\theta) \]
However, we know that the force needed to overcome the weight component down the incline is still 60 N. Thus, the horizontal force must also be equal to 60 N to maintain equilibrium.

3. **Calculate the work done**:
The work done (W) is given by:
\[ W = F_{\text{parallel}} \times d \]
Since the effective force parallel to the incline is still 60 N:
\[ W = 60 \times 2 = 120 \text{ J} \]

### Final Answers:
- Work done when the force is parallel to the incline: **120 J**
- Work done when the force is in the horizontal direction: **120 J**