The heavier block in an atwood machine has a mass twice that of the lighter one. The tension in the string is 16.0 N when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest.

Explanation

To solve the problem of finding the decrease in gravitational potential energy during the first second after the Atwood machine system is released from rest, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses**:
Let the mass of the lighter block be \( m \). Therefore, the mass of the heavier block will be \( 2m \).

2. **Write the Forces Acting on Each Block**:
- For the heavier block (mass \( 2m \)):
\[ F_{\text{net, heavy}} = 2mg - T = 2ma \]
- For the lighter block (mass \( m \)):
\[ F_{\text{net, light}} = T - mg = ma \]

3. **Combine the Equations**:
Adding the two equations:
\[ (2mg - T) + (T - mg) = 2ma + ma \]
This simplifies to:
\[ mg = 3ma \]
Therefore, we can express the acceleration \( a \) as:
\[ a = \frac{g}{3} \]

4. **Use the Given Tension**:
The tension \( T \) in the string is given as \( 16.0 \, \text{N} \). We can use this to find \( mg \):
From the equation for the lighter block:
\[ T - mg = ma \]
Substituting \( a \) from above:
\[ 16 - mg = m \left(\frac{g}{3}\right) \]
Rearranging gives:
\[ mg + \frac{mg}{3} = 16 \]
This simplifies to:
\[ \frac{4mg}{3} = 16 \]
Therefore:
\[ mg = 12 \, \text{N} \]

5. **Calculate the Displacement**:
The displacement \( S \) of the heavier block during the first second can be calculated using the second equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \) and \( t = 1 \, \text{s} \):
\[ S = 0 + \frac{1}{2} \left(\frac{g}{3}\right) (1^2) = \frac{g}{6} \]

6. **Calculate the Decrease in Potential Energy**:
The decrease in gravitational potential energy \( \Delta PE \) for the heavier block (which moves downward) is given by:
\[ \Delta PE = \text{Weight} \times \text{Displacement} = (2m \cdot g) \cdot S \]
Substituting \( mg = 12 \, \text{N} \) gives:
\[ \Delta PE = 2 \cdot 12 \cdot \frac{g}{6} \]
Since \( g \approx 10 \, \text{m/s}^2 \):
\[ \Delta PE = 2 \cdot 12 \cdot \frac{10}{6} = 20 \, \text{J} \]

### Final Answer:
The decrease in gravitational potential energy during the first second after the system is released from rest is **20 Joules**.