A block of mass m is hung vertically from an elastic thread of force constant `mg//a`. Initially the thread was at its natural length and the block is allowed to fall freely. Kinetic energy of the block when it passes through the equilibrium position will be

Options

A. mga
B. `(mga)/(2)`
C. zero
D. 2mga

Answer

To solve the problem, we need to determine the kinetic energy of a block of mass \( m \) when it passes through the equilibrium position after being released from rest. The elastic thread has a force constant given by \( k = \frac{mg}{a} \).

### Step-by-Step Solution:

1. **Identify the Equilibrium Position**:
The equilibrium position occurs when the gravitational force acting on the block is balanced by the elastic force of the thread. At this position, we have:
\[ mg = k \cdot x_0 \]
where \( x_0 \) is the extension of the thread at equilibrium.

2. **Substituting the Value of \( k \)**:
We know from the problem that \( k = \frac{mg}{a} \). Substituting this into the equilibrium equation gives:
\[ mg = \left(\frac{mg}{a}\right) x_0 \]

3. **Solving for \( x_0 \)**:
Rearranging the equation to find \( x_0 \):
\[ x_0 = a \]

4. **Calculating Work Done by Gravity**:
When the block falls from its initial position to the equilibrium position, the work done by gravity (\( W_g \)) is given by:
\[ W_g = mg \cdot x_0 = mg \cdot a \]

5. **Calculating the Work Done on the Thread**:
The work done on the elastic thread when it stretches from its natural length to the equilibrium position is given by:
\[ W_{\text{thread}} = \frac{1}{2} k x_0^2 \]
Substituting \( k \) and \( x_0 \):
\[ W_{\text{thread}} = \frac{1}{2} \left(\frac{mg}{a}\right) a^2 = \frac{1}{2} m g a \]

6. **Applying the Work-Energy Principle**:
The total work done on the block is equal to the change in kinetic energy (\( KE \)) as it passes through the equilibrium position:
\[ W_g - W_{\text{thread}} = KE \]
Substituting the values we found:
\[ mg a - \frac{1}{2} mg a = KE \]
Simplifying this gives:
\[ KE = \frac{1}{2} mg a \]

### Final Answer:
The kinetic energy of the block when it passes through the equilibrium position is:
\[ KE = \frac{1}{2} mg a \]