A 75 kg man stands in a lift. What force does the floor does the floor exert on him when the elevator starts moving upwards with an acceleration of `2.0ms^(-2)`. ? Take g = 10`ms^(-2)`.

Answer

To find the force that the floor exerts on the man when the elevator starts moving upwards with an acceleration of \(2.0 \, \text{m/s}^2\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the man, \(m = 75 \, \text{kg}\)
- Acceleration of the lift, \(a = 2.0 \, \text{m/s}^2\)
- Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\)

2. **Calculate the Weight of the Man:**
The weight of the man (\(W\)) can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the values:
\[ W = 75 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 750 \, \text{N} \]

3. **Calculate the Pseudo Force:**
When the lift accelerates upwards, a pseudo force acts downwards on the man. This force (\(F_{\text{pseudo}}\)) can be calculated using:
\[ F_{\text{pseudo}} = m \cdot a \]
Substituting the values:
\[ F_{\text{pseudo}} = 75 \, \text{kg} \cdot 2.0 \, \text{m/s}^2 = 150 \, \text{N} \]

4. **Calculate the Normal Force Exerted by the Floor:**
The normal force (\(N\)) exerted by the floor on the man must balance both the weight of the man and the pseudo force. Therefore, we can express this as:
\[ N = W + F_{\text{pseudo}} \]
Substituting the values:
\[ N = 750 \, \text{N} + 150 \, \text{N} = 900 \, \text{N} \]

5. **Convert the Normal Force to kgf:**
To convert the normal force from Newtons to kilograms-force (kgf), we can use the relation:
\[ 1 \, \text{kgf} = 10 \, \text{N} \]
Thus,
\[ N_{\text{kgf}} = \frac{N}{g} = \frac{900 \, \text{N}}{10 \, \text{m/s}^2} = 90 \, \text{kgf} \]

### Final Answer:
The force that the floor exerts on the man when the elevator starts moving upwards with an acceleration of \(2.0 \, \text{m/s}^2\) is \(900 \, \text{N}\) or \(90 \, \text{kgf}\).