A 0.50 kg object moves in a horizontal circular track of radius of 2.5 m. An external force of 3.0 N, always tangent to the track, causes the object to speed up as it goes around. The work done by the external force as the object makes one revolution is :

Options

A. 24 J
B. 47 J
C. 59 J
D. 94 J

Explanation

To find the work done by the external force as the object makes one revolution, we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done (W) by a force is given by the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
where:
- \( F \) is the magnitude of the force,
- \( d \) is the displacement,
- \( \theta \) is the angle between the force and the direction of displacement.

### Step 2: Identify the Given Values
From the problem:
- Mass of the object, \( m = 0.50 \, \text{kg} \) (not directly needed for work calculation),
- Radius of the circular track, \( r = 2.5 \, \text{m} \),
- External force, \( F = 3.0 \, \text{N} \).

### Step 3: Calculate the Displacement for One Revolution
The displacement \( d \) for one complete revolution around a circular track is the circumference of the circle, given by:
\[ d = 2 \pi r \]
Substituting the value of \( r \):
\[ d = 2 \pi (2.5) \]

### Step 4: Calculate the Circumference
Now, calculate the circumference:
\[ d = 2 \times 3.14 \times 2.5 \]
\[ d = 15.7 \, \text{m} \]

### Step 5: Determine the Angle
Since the force is always tangent to the track, the angle \( \theta \) between the force and the displacement is \( 0^\circ \). Therefore, \( \cos(0^\circ) = 1 \).

### Step 6: Calculate the Work Done
Now, substitute the values into the work done formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
\[ W = 3.0 \, \text{N} \cdot 15.7 \, \text{m} \cdot 1 \]
\[ W = 47.1 \, \text{J} \]

### Conclusion
The work done by the external force as the object makes one revolution is approximately \( 47.1 \, \text{J} \).

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