A lift carries a person with a weighting machine in it, up and then down with the same constant acceleration. The weights recorded by the machine in the two cases are in the ratio 5:3. The acceleation of the lift is

Options

A. ` 12.45ms^(-2)`
B. ` 3.8ms^(-2)`
C. ` 2.5ms^(-2)`
D. ` 3.67ms^(-2)`

Answer

To solve the problem, we need to analyze the forces acting on the person in the lift when it moves upwards and downwards with constant acceleration.

### Step-by-Step Solution:

1. **Identify the Forces When the Lift Moves Upwards:**
- When the lift moves upwards with acceleration \( a \), the forces acting on the person are:
- Weight of the person: \( mg \) (downward)
- Normal force from the weighing machine: \( N \) (upward)
- Pseudo force due to acceleration of the lift: \( ma \) (downward)
- According to Newton's second law, the net force is given by:
\[ N - mg - ma = 0 \]
- Rearranging gives:
\[ N = mg + ma \]

2. **Identify the Forces When the Lift Moves Downwards:**
- When the lift moves downwards with the same acceleration \( a \), the forces acting on the person are:
- Weight of the person: \( mg \) (downward)
- Normal force from the weighing machine: \( N' \) (upward)
- Pseudo force due to acceleration of the lift: \( ma \) (upward)
- According to Newton's second law, the net force is given by:
\[ N' - mg + ma = 0 \]
- Rearranging gives:
\[ N' = mg - ma \]

3. **Set Up the Ratio of the Weights:**
- We are given that the readings of the weighing machine in the two cases are in the ratio \( \frac{5}{3} \):
\[ \frac{N}{N'} = \frac{5}{3} \]
- Substituting the expressions for \( N \) and \( N' \):
\[ \frac{mg + ma}{mg - ma} = \frac{5}{3} \]

4. **Cross Multiply to Solve for \( a \):**
- Cross multiplying gives:
\[ 3(mg + ma) = 5(mg - ma) \]
- Expanding both sides:
\[ 3mg + 3ma = 5mg - 5ma \]
- Rearranging terms:
\[ 3ma + 5ma = 5mg - 3mg \]
\[ 8ma = 2mg \]

5. **Solve for the Acceleration \( a \):**
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ 8a = 2g \]
- Thus, we find:
\[ a = \frac{g}{4} \]

6. **Substituting the Value of \( g \):**
- Assuming \( g \approx 10 \, \text{m/s}^2 \):
\[ a = \frac{10}{4} = 2.5 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the lift is \( a = \frac{g}{4} \) or approximately \( 2.5 \, \text{m/s}^2 \).

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