A man of weight W is standing on a lift which is moving upward with an acceleration a. The apparent weight of the man is

Options

A. `W(1+a/g)`
B. `W`
C. `W(1-a/g)`
D. `W(1-(a^(2))/(g^(2)))`

Answer

To find the apparent weight of a man standing in a lift that is moving upward with an acceleration \( a \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Man
When the lift is at rest, the only force acting on the man is his weight \( W \), which can be expressed as:
\[ W = mg \]
where \( m \) is the mass of the man and \( g \) is the acceleration due to gravity.

### Step 2: Identify the Pseudo Force
When the lift accelerates upwards with an acceleration \( a \), the man experiences a pseudo force acting downwards. The magnitude of this pseudo force \( F_{\text{pseudo}} \) is given by:
\[ F_{\text{pseudo}} = ma \]

### Step 3: Calculate the Apparent Weight
The apparent weight \( W' \) of the man is the sum of his actual weight and the pseudo force acting on him. Therefore, we can express the apparent weight as:
\[ W' = W + F_{\text{pseudo}} = mg + ma \]

### Step 4: Factor Out the Mass
We can factor out the mass \( m \) from the equation:
\[ W' = m(g + a) \]

### Step 5: Substitute for Weight
Since \( W = mg \), we can substitute \( mg \) with \( W \):
\[ W' = W + \frac{W}{g} \cdot a \]
Rearranging gives:
\[ W' = W \left(1 + \frac{a}{g}\right) \]

### Conclusion
Thus, the apparent weight of the man when the lift is moving upward with an acceleration \( a \) is:
\[ W' = W \left(1 + \frac{a}{g}\right) \]