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. (a)24 J
B. (b)47 J
C. (c)59 J
D. (d)94 J

Explanation

To solve the problem of calculating the work done by the external force on a 0.50 kg object moving in a horizontal circular track of radius 2.5 m, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the object (m) = 0.50 kg
- Radius of the circular track (r) = 2.5 m
- External force (F) = 3.0 N (tangential to the track)

2. **Calculate the Circumference of the Circular Track:**
The distance traveled by the object in one complete revolution (circumference) can be calculated using the formula:
\[ \text{Circumference} = 2\pi r \]
Substituting the value of r:
\[ \text{Circumference} = 2 \times \pi \times 2.5 \]
\[ \text{Circumference} \approx 2 \times \frac{22}{7} \times 2.5 \approx 15.71 \text{ m} \]

3. **Calculate the Work Done by the External Force:**
The work done (W) by the external force when the object makes one complete revolution can be calculated using the formula:
\[ W = F \times d \]
where \(d\) is the distance traveled (circumference). Thus,
\[ W = 3.0 \, \text{N} \times 15.71 \, \text{m} \]
\[ W \approx 47.13 \, \text{J} \]

4. **Final Result:**
The work done by the external force as the object makes one revolution is approximately:
\[ W \approx 47 \, \text{J} \]

### Conclusion:
The work done by the external force is approximately 47 Joules.

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