In the arrangement shown in figure in which four blocks are suspended by cords . The top card loops over a frictionless pulley and pulls with a force of magnitude 98 N on the wall to which is attached . If the tensions in the shorter cords are ` T_(1) = 58.8 N, T_(2) = 49. 0 and T_(3) = 9.8 N,`

Answer

To solve the problem, we need to analyze the forces acting on each block and use the given tensions to find the masses of the blocks A, B, C, and D.

### Step-by-Step Solution:

1. **Understanding the Forces:**
- The system consists of four blocks (A, B, C, D) suspended by cords over a frictionless pulley.
- The top cord pulls with a force of 98 N.
- The tensions in the cords are given as:
- \( T_1 = 58.8 \, \text{N} \)
- \( T_2 = 49.0 \, \text{N} \)
- \( T_3 = 9.8 \, \text{N} \)

2. **Finding the Mass of Block D:**
- The tension \( T_3 \) is the force acting on block D (the lowest block).
- Using the formula for tension, \( T = mg \), where \( g = 9.8 \, \text{m/s}^2 \):
\[ T_3 = m_D \cdot g \implies 9.8 = m_D \cdot 9.8 \]
- Solving for \( m_D \):
\[ m_D = 1.0 \, \text{kg} \]

3. **Finding the Mass of Block C:**
- The tension \( T_2 \) supports both block C and block D.
- The equation for block C becomes:
\[ T_2 = m_C \cdot g + T_3 \]
- Substituting the known values:
\[ 49.0 = m_C \cdot 9.8 + 9.8 \]
- Rearranging gives:
\[ 49.0 - 9.8 = m_C \cdot 9.8 \implies 39.2 = m_C \cdot 9.8 \]
- Solving for \( m_C \):
\[ m_C = \frac{39.2}{9.8} = 4.0 \, \text{kg} \]

4. **Finding the Mass of Block B:**
- The tension \( T_1 \) supports blocks A and B, as well as the weight of block C.
- The equation for block B becomes:
\[ T_1 = m_B \cdot g + T_2 \]
- Substituting the known values:
\[ 58.8 = m_B \cdot 9.8 + 49.0 \]
- Rearranging gives:
\[ 58.8 - 49.0 = m_B \cdot 9.8 \implies 9.8 = m_B \cdot 9.8 \]
- Solving for \( m_B \):
\[ m_B = 1.0 \, \text{kg} \]

5. **Finding the Mass of Block A:**
- The total weight supported by the top cord (which is equal to the pulling force of 98 N) is the sum of the weights of blocks A, B, and C:
- The equation becomes:
\[ 98 = m_A \cdot g + T_1 \]
- Substituting the known values:
\[ 98 = m_A \cdot 9.8 + 58.8 \]
- Rearranging gives:
\[ 98 - 58.8 = m_A \cdot 9.8 \implies 39.2 = m_A \cdot 9.8 \]
- Solving for \( m_A \):
\[ m_A = \frac{39.2}{9.8} = 4.0 \, \text{kg} \]

### Summary of Masses:
- \( m_A = 4.0 \, \text{kg} \)
- \( m_B = 1.0 \, \text{kg} \)
- \( m_C = 4.0 \, \text{kg} \)
- \( m_D = 1.0 \, \text{kg} \)