A man, of mass 60kg, is riding in a lift. The weights of the man, when the lift is acceleration upwards and downwards at `2ms^(-2)` are respectively.
(Taking `g=10ms^(-2)`)

Options

A. 720 N and 480N
B. 480N and 720N
C. 600N and 600N
D. None of these

Explanation

To solve the problem, we need to calculate the apparent weight of a man in a lift that is accelerating both upwards and downwards. The apparent weight is given by the normal force exerted by the weighing machine, which can be calculated using Newton's second law of motion.

### Given:
- Mass of the man, \( m = 60 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Acceleration of the lift, \( a = 2 \, \text{m/s}^2 \)

### Step 1: Calculate the weight of the man
The weight of the man (force due to gravity) is given by:
\[ W = m \cdot g = 60 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 600 \, \text{N} \]

### Step 2: Case 1 - Lift accelerating upwards
When the lift accelerates upwards, the apparent weight (normal force \( N \)) can be calculated using the equation:
\[ N = W + m \cdot a \]
Substituting the values:
\[ N = 600 \, \text{N} + (60 \, \text{kg} \cdot 2 \, \text{m/s}^2) = 600 \, \text{N} + 120 \, \text{N} = 720 \, \text{N} \]

### Step 3: Case 2 - Lift accelerating downwards
When the lift accelerates downwards, the apparent weight can be calculated using the equation:
\[ N = W - m \cdot a \]
Substituting the values:
\[ N = 600 \, \text{N} - (60 \, \text{kg} \cdot 2 \, \text{m/s}^2) = 600 \, \text{N} - 120 \, \text{N} = 480 \, \text{N} \]

### Final Results:
- Apparent weight when the lift accelerates upwards: **720 N**
- Apparent weight when the lift accelerates downwards: **480 N**

### Summary:
- When the lift accelerates upwards, the man feels heavier, and his weight is 720 N.
- When the lift accelerates downwards, the man feels lighter, and his weight is 480 N.