To solve the problem, we need to determine the weight of a person inside a lift that suddenly breaks down. The person has a mass of 60 kg and is initially descending with an acceleration of 2 m/s².
### Step-by-Step Solution:
1. **Understanding Weight and Apparent Weight**:
- The weight of an object is given by the formula:
\[
W = mg
\]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).
- The apparent weight (normal force, \( N \)) experienced by the person in the lift is affected by the acceleration of the lift.
2. **Calculating the Weight of the Person**:
- The actual weight of the person is:
\[
W = 60 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 588 \, \text{N}
\]
3. **Setting Up the Free Body Diagram**:
- When the lift is descending with an acceleration of \( 2 \, \text{m/s}^2 \), we can analyze the forces acting on the person:
- Downward force: \( mg \) (weight)
- Upward force: \( N \) (normal force or apparent weight)
4. **Applying Newton's Second Law**:
- According to Newton's second law, the net force acting on the person can be written as:
\[
F_{\text{net}} = ma
\]
- Taking downward as positive, we have:
\[
mg - N = ma
\]
- Rearranging gives:
\[
N = mg - ma
\]
5. **Substituting Values**:
- Substitute \( m = 60 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( a = 2 \, \text{m/s}^2 \):
\[
N = (60 \, \text{kg} \times 9.8 \, \text{m/s}^2) - (60 \, \text{kg} \times 2 \, \text{m/s}^2)
\]
\[
N = 588 \, \text{N} - 120 \, \text{N} = 468 \, \text{N}
\]
6. **Condition When the Lift Breaks**:
- When the lift suddenly breaks, it is in free fall. In this case, the acceleration of the lift becomes equal to the acceleration due to gravity (\( g \)).
- Therefore, the apparent weight becomes:
\[
N = mg - mg = 0
\]
### Final Answer:
- The weight of the person inside the lift when it suddenly breaks is **0 N**. This indicates that the person experiences weightlessness.
