A block of mass `m` attached in the lower and vertical spring The spring is hung from a calling and force constant value `k` The mass is released from rest with the spring initially unstreched The maximum value of extension produced in the length of the spring will be

Options

A. `(2Mg)/(k)`
B. `(4Mg)/(k)`
C. `(Mg)/(2k)`
D. `(Mg)/(k)`

Answer

To solve the problem of finding the maximum extension produced in the length of a spring when a block of mass \( m \) is attached to it, we can use the work-energy theorem. Here’s a step-by-step solution:

### Step 1: Understand the System
We have a vertical spring with a spring constant \( k \) and a mass \( m \) attached to its lower end. The spring is initially unstretched, and the mass is released from rest.

### Step 2: Identify Key Points
When the mass is released, it will fall under the influence of gravity until it reaches the maximum extension of the spring. At this point, the velocity of the mass will be zero.

### Step 3: Apply the Work-Energy Theorem
According to the work-energy theorem, the work done on the system is equal to the change in kinetic energy. Initially, the kinetic energy is zero because the mass is at rest.

The work done by the gravitational force when the mass falls a distance \( L_{max} \) is:
\[ W_{gravity} = mgh = mgL_{max} \]
where \( h \) is the distance fallen, which is equal to the maximum extension \( L_{max} \).

The work done by the spring force when it is stretched by \( L_{max} \) is:
\[ W_{spring} = -\frac{1}{2} k L_{max}^2 \]
(The negative sign indicates that the spring force acts in the opposite direction to the displacement.)

### Step 4: Set Up the Equation
At maximum extension, the total work done by the gravitational force equals the work done on the spring:
\[ mgL_{max} = \frac{1}{2} k L_{max}^2 \]

### Step 5: Rearrange the Equation
Rearranging the equation gives:
\[ \frac{1}{2} k L_{max}^2 - mgL_{max} = 0 \]
Factoring out \( L_{max} \):
\[ L_{max} \left( \frac{1}{2} k L_{max} - mg \right) = 0 \]
This gives us two solutions:
1. \( L_{max} = 0 \) (the trivial solution)
2. \( \frac{1}{2} k L_{max} - mg = 0 \)

### Step 6: Solve for Maximum Extension
From the second equation:
\[ \frac{1}{2} k L_{max} = mg \]
\[ L_{max} = \frac{2mg}{k} \]

### Conclusion
The maximum extension produced in the length of the spring is:
\[ L_{max} = \frac{2mg}{k} \]