A thin uniform steel chain is 10 meters long with a mass density of 2 kilograms per meter. One end of the chain is attached to a horizontal axis having radius that is small compared to the legnth of the chain. If the chain initially hangs vertically, the work required to slowly wind it up on to the axle is closest to

Options

A. 100 J
B. 200 J
C. 1000 J
D. 2000 J

Explanation

To solve the problem of calculating the work required to wind up a thin uniform steel chain of length 10 meters and mass density of 2 kg/m, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the total mass of the chain:**
The mass density (λ) of the chain is given as 2 kg/m. The total length (L) of the chain is 10 m. Therefore, the total mass (M) of the chain can be calculated as:
\[ M = \lambda \times L = 2 \, \text{kg/m} \times 10 \, \text{m} = 20 \, \text{kg} \]

2. **Identify the small segment of the chain:**
Consider a small segment of the chain of length \(dy\) at a height \(y\) from the bottom of the chain. The mass of this small segment (\(dm\)) can be expressed as:
\[ dm = \lambda \times dy = 2 \, \text{kg/m} \times dy \]

3. **Calculate the weight of the small segment:**
The weight (\(dW\)) of this small segment is given by:
\[ dW = dm \times g = (2 \, dy) \times g \]
where \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)).

4. **Determine the work done to lift this small segment:**
The work done to lift this small segment from height \(y\) to the horizontal axis (which is at height 0) is:
\[ dW = (2 \, dy) \times g \times y \]

5. **Integrate to find the total work done:**
To find the total work done to lift the entire chain, we integrate \(dW\) from \(y = 0\) to \(y = 10\):
\[ W = \int_0^{10} (2g \cdot y) \, dy \]
\[ W = 2g \int_0^{10} y \, dy \]
\[ W = 2g \left[ \frac{y^2}{2} \right]_0^{10} \]
\[ W = 2g \left[ \frac{10^2}{2} - 0 \right] = 2g \cdot 50 = 100g \]

6. **Substitute the value of \(g\):**
Using \(g \approx 9.81 \, \text{m/s}^2\):
\[ W = 100 \times 9.81 = 981 \, \text{J} \]

7. **Final result:**
The work required to slowly wind up the chain onto the axle is approximately:
\[ W \approx 981 \, \text{J} \]

### Conclusion:
The closest answer to the work required to slowly wind up the chain is approximately **981 Joules**.