n indentical blocks each of mass m are connected by means mass less strings and a pulling force F is applied on the first block. Find the tension in the string connecting `k^("th")` and ) `(k+1)^("th")` blocks. If another pulling force F is applied on the last block, find the tension in the same . String.

Options

A. `[F(n-k)//n]N,F`
B. `[F(n-k)//n]N,nF`
C. `[F(n+k)//n]N,F`
D. `[2F(n-k)n]N,F`

Answer

To solve the problem, we will break it down into two parts:

1. Finding the tension in the string connecting the \(k^{th}\) and \((k+1)^{th}\) blocks when a force \(F\) is applied on the first block.
2. Finding the tension in the same string when another force \(F\) is applied on the last block.

### Part 1: Tension between the \(k^{th}\) and \((k+1)^{th}\) blocks

**Step 1: Understand the system**
- We have \(n\) identical blocks, each with mass \(m\), connected by massless strings.
- A pulling force \(F\) is applied to the first block.

**Step 2: Calculate the total mass being accelerated**
- The total mass being accelerated by the force \(F\) is \(n \cdot m\).

**Step 3: Calculate the acceleration of the system**
- Using Newton's second law, \(F = ma\), we can find the acceleration \(a\):
\[ a = \frac{F}{n \cdot m} \]

**Step 4: Analyze the forces on the \(k^{th}\) block**
- The \(k^{th}\) block is being pulled by the tension \(T_k\) from the string connecting it to the \((k-1)^{th}\) block and is pulling the \((k+1)^{th}\) block with tension \(T_{k+1}\).
- The net force on the \(k^{th}\) block can be expressed as:
\[ T_k - T_{k+1} = m \cdot a \]

**Step 5: Express the tension in terms of \(F\)**
- The tension \(T_k\) can be calculated as:
\[ T_k = F - (n - k) \cdot m \cdot a \]
- Substituting \(a\) from Step 3:
\[ T_k = F - (n - k) \cdot m \cdot \frac{F}{n \cdot m} \]
\[ T_k = F - \frac{(n - k) \cdot F}{n} \]
\[ T_k = F \left(1 - \frac{n - k}{n}\right) \]
\[ T_k = F \left(\frac{k}{n}\right) \]

### Part 2: Tension when a force \(F\) is applied on the last block

**Step 6: Analyze the new situation**
- Now, a force \(F\) is applied to the last block (the \(n^{th}\) block).
- The forces on the blocks are now equal and opposite.

**Step 7: Calculate the tension in the string**
- Since both ends are being pulled with the same force \(F\), the tension in the string connecting the \(k^{th}\) and \((k+1)^{th}\) blocks will be equal to \(F\) because the forces are balanced and the system is in equilibrium.

### Final Tensions
- The tension in the string connecting the \(k^{th}\) and \((k+1)^{th}\) blocks when a force \(F\) is applied on the first block is:
\[ T_k = \frac{k}{n} F \]
- The tension in the same string when a force \(F\) is applied on the last block is:
\[ T_k = F \]