A man drags a block through 10 m on rough surface (`mu`=0.5).A force of `sqrt(3)` kN acting at `30^(@)` to the horizontal .The work done by applied force is

Options

A. zero
B. 15 kJ
C. 5kJ
D. 10 kJ

Answer

To find the work done by the applied force when dragging a block through a rough surface, we can follow these steps:

### Step 1: Identify the given values
- Displacement (d) = 10 m
- Coefficient of friction (μ) = 0.5 (not needed for work done calculation)
- Applied force (F) = √3 kN = √3 × 1000 N = 1732 N (since 1 kN = 1000 N)
- Angle (θ) = 30° (angle between the applied force and the horizontal)

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

### Step 3: Substitute the values into the formula
Now, we can substitute the known values into the formula:
\[ W = (1732 \, \text{N}) \cdot (10 \, \text{m}) \cdot \cos(30°) \]

### Step 4: Calculate cos(30°)
We know that:
\[ \cos(30°) = \frac{\sqrt{3}}{2} \]

### Step 5: Substitute cos(30°) into the equation
Now we can substitute cos(30°) into the equation:
\[ W = (1732 \, \text{N}) \cdot (10 \, \text{m}) \cdot \left(\frac{\sqrt{3}}{2}\right) \]

### Step 6: Calculate the work done
Now we can calculate the work done:
\[ W = 17320 \cdot \frac{\sqrt{3}}{2} \]
\[ W = 8660 \sqrt{3} \, \text{J} \]

### Step 7: Convert to kilojoules
Since 1 kJ = 1000 J, we convert the work done to kilojoules:
\[ W = \frac{8660 \sqrt{3}}{1000} \, \text{kJ} \]
\[ W \approx 15 \, \text{kJ} \] (since \(\sqrt{3} \approx 1.732\))

### Final Answer
The work done by the applied force is approximately **15 kJ**.

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