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 applied force is

Options

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

Answer

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

### Step 1: Identify the given values
- Mass of the block, \( m = 2.5 \, \text{kg} \) (not needed for work calculation)
- Displacement, \( s = 2.2 \, \text{m} \)
- Applied force, \( F = 16 \, \text{N} \)
- Angle of the force above the horizontal, \( \theta = 45^\circ \)

### Step 2: Write the formula for work done
The work done \( W \) by a force is given by the formula:
\[ W = F \cdot s \cdot \cos(\theta) \]
where:
- \( F \) is the magnitude of the force,
- \( s \) is the displacement,
- \( \theta \) is the angle between the force and the direction of displacement.

### Step 3: Calculate \( \cos(45^\circ) \)
We know that:
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.7071 \]

### Step 4: Substitute the values into the formula
Now we can substitute the values into the work done formula:
\[ W = 16 \, \text{N} \cdot 2.2 \, \text{m} \cdot \cos(45^\circ) \]
\[ W = 16 \cdot 2.2 \cdot \frac{1}{\sqrt{2}} \]

### Step 5: Calculate the work done
Calculating the above expression:
\[ W = 16 \cdot 2.2 \cdot \frac{1}{\sqrt{2}} \approx 16 \cdot 2.2 \cdot 0.7071 \]
\[ W \approx 16 \cdot 2.2 \cdot 0.7071 \approx 24.89 \, \text{J} \]

### Step 6: Round off the answer
Rounding off the answer, we get:
\[ W \approx 24.9 \, \text{J} \]

### Final Answer
The work done by the applied force is approximately **24.9 Joules**.

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