A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49N, when the lift is stationary. If the lift moves downward with an acceleration of `5m//2^2`, the reading of the spring balance will be

Options

A. 49 N
B. 24 N
C. 74 N
D. 15 N

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial condition
When the lift is stationary, the reading of the spring balance is equal to the weight of the bag. The weight (W) of the bag can be expressed as:
\[ W = mg \]
where \( m \) is the mass of the bag and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 2: Calculate the mass of the bag
Given that the spring balance reads \( 49 \, \text{N} \) when the lift is stationary, we can set up the equation:
\[ mg = 49 \, \text{N} \]
Now, we can solve for \( m \):
\[ m = \frac{49}{g} = \frac{49}{9.8} \]
Calculating this gives:
\[ m = 5 \, \text{kg} \]

### Step 3: Analyze the situation when the lift is moving downward
When the lift is moving downward with an acceleration \( a = 5 \, \text{m/s}^2 \), the apparent weight (R') of the bag can be calculated using the formula:
\[ R' = mg - ma \]
Here, \( mg \) is the weight of the bag and \( ma \) is the force due to the acceleration of the lift.

### Step 4: Substitute the known values
We already calculated \( m = 5 \, \text{kg} \). Now substituting the values into the equation:
\[ R' = mg - ma \]
\[ R' = (5 \, \text{kg} \cdot 9.8 \, \text{m/s}^2) - (5 \, \text{kg} \cdot 5 \, \text{m/s}^2) \]
Calculating \( mg \):
\[ mg = 49 \, \text{N} \]
Calculating \( ma \):
\[ ma = 25 \, \text{N} \]
Now substituting these values back into the equation:
\[ R' = 49 \, \text{N} - 25 \, \text{N} \]
\[ R' = 24 \, \text{N} \]

### Step 5: Conclusion
The reading of the spring balance when the lift is moving downward with an acceleration of \( 5 \, \text{m/s}^2 \) is \( 24 \, \text{N} \).

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