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

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Understand the situation
When the lift is stationary, the spring balance reads 49 N. This reading represents the weight of the bag, which can be expressed as:
\[ W = mg \]
where \( W \) is the weight (49 N), \( m \) 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
Using the weight of the bag:
\[ mg = 49 \, \text{N} \]
We can rearrange this to find the mass \( m \):
\[ m = \frac{W}{g} = \frac{49}{9.8} \approx 5 \, \text{kg} \]

### Step 3: Analyze the situation when the lift is accelerating downwards
When the lift accelerates downwards with an acceleration \( a = 5 \, \text{m/s}^2 \), the effective weight of the bag (the reading on the spring balance) will change. The apparent weight \( F \) can be calculated using the formula:
\[ F = mg - ma \]
where \( ma \) is the force due to the downward acceleration of the lift.

### Step 4: Substitute the values into the equation
Now substituting the values we have:
- \( m = 5 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( a = 5 \, \text{m/s}^2 \)

We can calculate \( F \):
\[ F = mg - ma = (5 \times 9.8) - (5 \times 5) \]
\[ F = 49 - 25 = 24 \, \text{N} \]

### Conclusion
The reading on the spring balance when the lift is moving downwards with an acceleration of \( 5 \, \text{m/s}^2 \) will be:
\[ \boxed{24 \, \text{N}} \]