A block of mass 2.5 kg is pushed 2.2. m along a frictionless horizontal table by a constant 16 N force directed `45^(@)` above the horizontal. Then
Workdone by the force of gravity

Options

A. 24.9J
B. 34.9J
C. 2.49J
D. zero

Answer

To find the work done by the force of gravity on the block, we can follow these steps:

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
1. The applied force of 16 N directed at an angle of 45 degrees above the horizontal.
2. The gravitational force acting downward, which can be calculated using the formula \( F_g = mg \).

### Step 2: Calculate the gravitational force
Given:
- Mass of the block, \( m = 2.5 \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

The gravitational force can be calculated as:
\[ F_g = mg = 2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 24.5 \, \text{N} \]

### Step 3: Determine the displacement
The displacement of the block is given as:
\[ s = 2.2 \, \text{m} \]

### Step 4: Calculate the work done by the force of gravity
The work done by a force is given by the formula:
\[ W = F \cdot s \cdot \cos(\theta) \]
where:
- \( W \) is the work done,
- \( F \) is the force,
- \( s \) is the displacement,
- \( \theta \) is the angle between the force and the displacement.

In this case, the angle \( \theta \) between the gravitational force (which acts downward) and the displacement (which is horizontal) is \( 90^\circ \).

### Step 5: Substitute the values into the work formula
Since \( \cos(90^\circ) = 0 \):
\[ W = F_g \cdot s \cdot \cos(90^\circ) = 24.5 \, \text{N} \cdot 2.2 \, \text{m} \cdot 0 = 0 \]

### Conclusion
The work done by the force of gravity on the block is:
\[ W = 0 \, \text{J} \]