A 4.0 kg block is suspended from the ceiling of an elevator through a string having a linear mass desity of `19.2xx10^-3 kgm^-1`. Find the speed (with respect to the string) with which a wave pulse can proceed on the string if the elavator accelerates up at the rate of `2.0ms^-2. Take g=10 ms^-2`.

Explanation

To solve the problem, we need to find the speed of a wave pulse on a string that is subjected to tension due to a block suspended in an accelerating elevator. Here are the steps to find the solution:

### Step 1: Identify the forces acting on the block
The block has two forces acting on it:
- The gravitational force (weight) acting downwards, which is given by \( F_g = mg \).
- The tension \( T \) in the string acting upwards.

### Step 2: Calculate the weight of the block
Given:
- Mass of the block \( m = 4.0 \, \text{kg} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

The weight of the block is calculated as:
\[ F_g = mg = 4.0 \, \text{kg} \times 10 \, \text{m/s}^2 = 40 \, \text{N} \]

### Step 3: Apply Newton's second law to find the tension in the string
The elevator is accelerating upwards with an acceleration \( a = 2.0 \, \text{m/s}^2 \). According to Newton's second law:
\[ T - F_g = ma \]
Substituting the known values:
\[ T - 40 \, \text{N} = 4.0 \, \text{kg} \times 2.0 \, \text{m/s}^2 \]
\[ T - 40 = 8 \quad \Rightarrow \quad T = 40 + 8 = 48 \, \text{N} \]

### Step 4: Calculate the linear mass density of the string
The linear mass density \( \mu \) is given as:
\[ \mu = 19.2 \times 10^{-3} \, \text{kg/m} \]

### Step 5: Use the wave speed formula
The speed of a wave pulse on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values of tension and linear mass density:
\[ v = \sqrt{\frac{48 \, \text{N}}{19.2 \times 10^{-3} \, \text{kg/m}}} \]

### Step 6: Simplify the calculation
Calculating the denominator:
\[ 19.2 \times 10^{-3} = 0.0192 \, \text{kg/m} \]
Now substituting back:
\[ v = \sqrt{\frac{48}{0.0192}} = \sqrt{2500} = 50 \, \text{m/s} \]

### Final Answer
The speed with which a wave pulse can proceed on the string is:
\[ \boxed{50 \, \text{m/s}} \]