n identical cubes each of amss m an side I are on the horizontal surface. Then the minimum amount of work done to arrange then one on the ther is

Options

A. nmmgl
B. `(mgln^(2))/2`
C. `(mgln(n-1))/2`
D. `(mgln(n+1))/2`

Answer

To solve the problem of finding the minimum amount of work done to arrange \( n \) identical cubes, each of mass \( m \) and side length \( l \), one on top of the other, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have \( n \) identical cubes, each with mass \( m \) and side length \( l \), placed on a horizontal surface. We need to stack these cubes vertically.

2. **Calculating Work Done for Each Cube**:
- The first cube is already on the ground, so no work is needed to place it.
- The second cube needs to be lifted to a height of \( l \). The work done \( W_1 \) to lift it is given by:
\[ W_1 = m \cdot g \cdot l \]
- The third cube needs to be lifted to a height of \( 2l \). The work done \( W_2 \) is:
\[ W_2 = m \cdot g \cdot 2l \]
- Continuing this pattern, the \( k^{th} \) cube needs to be lifted to a height of \( (k-1)l \), where \( k \) ranges from 1 to \( n \).

3. **Total Work Done**:
- The total work done \( W \) to stack all \( n \) cubes is the sum of the work done for each cube:
\[ W = W_1 + W_2 + W_3 + \ldots + W_{n-1} \]
- This can be expressed as:
\[ W = m \cdot g \cdot l + m \cdot g \cdot 2l + m \cdot g \cdot 3l + \ldots + m \cdot g \cdot (n-1)l \]

4. **Factoring Out Common Terms**:
- We can factor out \( m \cdot g \cdot l \):
\[ W = m \cdot g \cdot l \cdot (1 + 2 + 3 + \ldots + (n-1)) \]

5. **Using the Formula for the Sum of First \( n \) Natural Numbers**:
- The sum of the first \( n-1 \) natural numbers is given by:
\[ S = \frac{(n-1)n}{2} \]
- Therefore, we can substitute this into our expression for \( W \):
\[ W = m \cdot g \cdot l \cdot \frac{(n-1)n}{2} \]

6. **Final Expression for Work Done**:
- Thus, the minimum amount of work done to arrange the cubes is:
\[ W = \frac{m \cdot g \cdot l \cdot (n-1)n}{2} \]

### Final Answer:
The minimum amount of work done to arrange \( n \) identical cubes, each of mass \( m \) and side \( l \), one on top of the other is:
\[ W = \frac{m \cdot g \cdot l \cdot (n-1)n}{2} \]