When a man holding spring balance in stationary lift then it's reading is `60 kg`. Now if lift starts descends with constant acceleration 1.8 m/`s^(2)` then what is the new reading of spring balance in newton (take g =10 m/`s^(2)`)

Answer

To solve the problem step by step, we need to analyze the forces acting on the man in the lift when it is descending with constant acceleration.

### Step 1: Understand the forces acting on the man in the lift
When the lift is stationary, the forces acting on the man are:
- The weight of the man (downward), which is given by \( mg \).
- The normal force (upward), which is the reading of the spring balance \( N \).

Since the lift is stationary, these forces are balanced:
\[ N = mg \]

### Step 2: Calculate the weight of the man
Given that the reading of the spring balance when stationary is \( 60 \, \text{kg} \), we can calculate the weight:
- \( m = 60 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)

Thus, the weight \( W \) is:
\[ W = mg = 60 \, \text{kg} \times 10 \, \text{m/s}^2 = 600 \, \text{N} \]

### Step 3: Analyze the situation when the lift is descending
When the lift descends with a constant acceleration \( a = 1.8 \, \text{m/s}^2 \), we need to find the new reading of the spring balance \( N' \).

In the accelerating frame of reference (the lift), the effective weight of the man is reduced because the lift is accelerating downwards. The net force acting on the man can be expressed as:
\[ N' = mg - ma \]

### Step 4: Substitute the values into the equation
Now, substituting the known values into the equation:
- \( m = 60 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( a = 1.8 \, \text{m/s}^2 \)

We get:
\[ N' = mg - ma \]
\[ N' = 60 \, \text{kg} \times 10 \, \text{m/s}^2 - 60 \, \text{kg} \times 1.8 \, \text{m/s}^2 \]
\[ N' = 600 \, \text{N} - 108 \, \text{N} \]
\[ N' = 492 \, \text{N} \]

### Final Answer
The new reading of the spring balance when the lift is descending with a constant acceleration of \( 1.8 \, \text{m/s}^2 \) is \( 492 \, \text{N} \).

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