Two blocks of masses 40 kg and 20 kg are connected by a wire that has a linear mass density of `1 g//m`. These blocks are being pulled across horizontal frictionless floor by horizontal force F that is applied to 20 kg block. A transverse wave travels on the wave between the blocks with a speed of `400 m//s` (relative to the wire). the lmass of the wire is negligible compared to the mass of the blocks. the magnitude of F is

Options

A. 160 N
B. 240 N
C. 320 N
D. 400 N

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the System
We have two blocks of masses \( m_1 = 40 \, \text{kg} \) and \( m_2 = 20 \, \text{kg} \) connected by a wire with a linear mass density \( \mu = 1 \, \text{g/m} = 0.001 \, \text{kg/m} \). A force \( F \) is applied to the 20 kg block, and a transverse wave travels along the wire with a speed of \( v = 400 \, \text{m/s} \).

**Hint:** Identify the masses and the properties of the wire to understand the system.

### Step 2: Calculate the Total Mass
The total mass of the system is the sum of the masses of the two blocks:
\[ m_{\text{total}} = m_1 + m_2 = 40 \, \text{kg} + 20 \, \text{kg} = 60 \, \text{kg} \]

**Hint:** The total mass is essential for calculating the acceleration of the system.

### Step 3: Relate Tension to Wave Speed
The speed of the wave in the wire is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire. Rearranging this gives:
\[ T = v^2 \cdot \mu \]

**Hint:** Use the wave speed formula to express tension in terms of wave speed and linear mass density.

### Step 4: Substitute Known Values
Substituting the known values into the tension equation:
\[ T = (400 \, \text{m/s})^2 \cdot (0.001 \, \text{kg/m}) = 160 \, \text{N} \]

**Hint:** Ensure units are consistent when performing calculations.

### Step 5: Relate Tension to Force
The tension in the wire is also related to the force \( F \) applied to the 20 kg block. The tension is what pulls the 40 kg block. The acceleration \( a \) of the system can be expressed as:
\[ a = \frac{F}{m_{\text{total}}} = \frac{F}{60} \]
The tension can also be expressed as:
\[ T = m_1 \cdot a = 40 \cdot \frac{F}{60} = \frac{2F}{3} \]

**Hint:** Use Newton's second law to relate force, mass, and acceleration.

### Step 6: Set Up the Equation
Now we have two expressions for tension:
1. \( T = 160 \, \text{N} \)
2. \( T = \frac{2F}{3} \)

Setting these equal gives:
\[ 160 = \frac{2F}{3} \]

**Hint:** Equate the two expressions for tension to find the force.

### Step 7: Solve for Force \( F \)
To find \( F \), multiply both sides by 3:
\[ 480 = 2F \]
Now divide by 2:
\[ F = 240 \, \text{N} \]

**Hint:** Isolate \( F \) to find its value.

### Final Answer
The magnitude of the force \( F \) applied to the 20 kg block is:
\[ \boxed{240 \, \text{N}} \]