An `80 kg` man is riding on a `40 kg` cart travelling at a speed of `2.5 m//s` on a frictionless horizontal plane jumps off the cart, such that, his velocity just after is zero with respect to ground. The work done by him on th esystem during his jump is given as `(A)/(4) KJ` (Ain integer). Find the value of A.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the system
We have a man with a mass of \( m_1 = 80 \, \text{kg} \) and a cart with a mass of \( m_2 = 40 \, \text{kg} \). The initial speed of the system (man + cart) is \( u = 2.5 \, \text{m/s} \).

### Step 2: Apply the conservation of momentum
Before the man jumps off, the total momentum of the system is given by:
\[ \text{Initial Momentum} = (m_1 + m_2) \cdot u = (80 + 40) \cdot 2.5 = 120 \cdot 2.5 = 300 \, \text{kg m/s} \]

After the man jumps off, his velocity with respect to the ground is \( 0 \, \text{m/s} \). Let the velocity of the cart after the man jumps off be \( v \).

The momentum after the jump is:
\[ \text{Final Momentum} = m_1 \cdot 0 + m_2 \cdot v = 0 + 40v = 40v \]

By conservation of momentum:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
\[ 300 = 40v \]
Solving for \( v \):
\[ v = \frac{300}{40} = 7.5 \, \text{m/s} \]

### Step 3: Calculate the change in kinetic energy
The change in kinetic energy (work done by the man on the system) can be calculated using the formula:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]

**Initial Kinetic Energy**:
\[ KE_{\text{initial}} = \frac{1}{2} (m_1 + m_2) u^2 = \frac{1}{2} \cdot 120 \cdot (2.5)^2 = \frac{1}{2} \cdot 120 \cdot 6.25 = 375 \, \text{J} \]

**Final Kinetic Energy**:
\[ KE_{\text{final}} = \frac{1}{2} m_2 v^2 = \frac{1}{2} \cdot 40 \cdot (7.5)^2 = \frac{1}{2} \cdot 40 \cdot 56.25 = 1125 \, \text{J} \]

### Step 4: Calculate the change in kinetic energy
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 1125 - 375 = 750 \, \text{J} \]

### Step 5: Convert Joules to Kilojoules
Since \( 1 \, \text{KJ} = 1000 \, \text{J} \):
\[ \Delta KE = 750 \, \text{J} = 0.75 \, \text{KJ} \]

### Step 6: Find the value of \( A \)
According to the problem, the work done by the man is given as \( \frac{A}{4} \, \text{KJ} \):
\[ \frac{A}{4} = 0.75 \implies A = 0.75 \times 4 = 3 \]

Thus, the value of \( A \) is \( 3 \).

### Final Answer
\[ \boxed{3} \]