A `60kg` man is inside a lift which is moving up with an acceleration of `2.45_(ms^(-2))`. The appar-ent percentage change in his weight is .

Options

A. `20%`
B. `25%`
C. `50%`
D. `75%`

Explanation

To solve the problem of finding the apparent percentage change in the weight of a 60 kg man inside a lift moving upwards with an acceleration of 2.45 m/s², we can follow these steps:

### Step 1: Calculate the actual weight of the man
The actual weight (W₁) of the man can be calculated using the formula:
\[ W₁ = m \cdot g \]
where:
- \( m = 60 \, \text{kg} \) (mass of the man)
- \( g = 9.8 \, \text{m/s}² \) (acceleration due to gravity)

Substituting the values:
\[ W₁ = 60 \, \text{kg} \cdot 9.8 \, \text{m/s}² = 588 \, \text{N} \]

### Step 2: Calculate the apparent weight of the man in the lift
When the lift is accelerating upwards, the apparent weight (W₂) can be calculated using the formula:
\[ W₂ = m \cdot (g + a) \]
where:
- \( a = 2.45 \, \text{m/s}² \) (acceleration of the lift)

Substituting the values:
\[ W₂ = 60 \, \text{kg} \cdot (9.8 \, \text{m/s}² + 2.45 \, \text{m/s}²) \]
\[ W₂ = 60 \, \text{kg} \cdot 12.25 \, \text{m/s}² = 735 \, \text{N} \]

### Step 3: Calculate the change in weight
The change in weight can be calculated as:
\[ \Delta W = W₂ - W₁ \]
Substituting the values:
\[ \Delta W = 735 \, \text{N} - 588 \, \text{N} = 147 \, \text{N} \]

### Step 4: Calculate the percentage change in weight
The percentage change in weight can be calculated using the formula:
\[ \text{Percentage Change} = \left( \frac{\Delta W}{W₁} \right) \cdot 100\% \]
Substituting the values:
\[ \text{Percentage Change} = \left( \frac{147 \, \text{N}}{588 \, \text{N}} \right) \cdot 100\% \]
\[ \text{Percentage Change} = 0.25 \cdot 100\% = 25\% \]

### Final Answer
The apparent percentage change in the man's weight is **25%**.

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