Across a horizontal floor, a 102 kg block is pulled at a constant speed of 5.5 m/s by an applied force of 125 N directed `38^(@)` above the horizontal. Calculate the rate at which the force does work on the block.

Answer

To calculate the rate at which the force does work on the block, we need to find the power exerted by the applied force. The power can be calculated using the formula:

\[ P = F \cdot v \cdot \cos(\theta) \]

Where:
- \( P \) is the power,
- \( F \) is the magnitude of the applied force,
- \( v \) is the velocity of the block,
- \( \theta \) is the angle between the force and the direction of motion.

### Step 1: Identify the given values
From the problem statement, we have:
- Mass of the block, \( m = 102 \, \text{kg} \) (not directly needed for power calculation)
- Applied force, \( F = 125 \, \text{N} \)
- Velocity, \( v = 5.5 \, \text{m/s} \)
- Angle, \( \theta = 38^\circ \)

### Step 2: Calculate the cosine of the angle
We need to find \( \cos(38^\circ) \). Using a calculator or trigonometric tables:

\[ \cos(38^\circ) \approx 0.788 \]

### Step 3: Substitute the values into the power formula
Now we can substitute the values into the power formula:

\[ P = F \cdot v \cdot \cos(\theta) \]

Substituting the known values:

\[ P = 125 \, \text{N} \cdot 5.5 \, \text{m/s} \cdot \cos(38^\circ) \]

\[ P = 125 \cdot 5.5 \cdot 0.788 \]

### Step 4: Calculate the power
Now we perform the multiplication:

\[ P = 125 \cdot 5.5 = 687.5 \]

Then multiply by \( \cos(38^\circ) \):

\[ P = 687.5 \cdot 0.788 \approx 542.75 \, \text{W} \]

### Final Answer
The rate at which the force does work on the block is approximately:

\[ P \approx 542.75 \, \text{W} \]