The problem describes a lift (elevator) with a total mass of 600 kg, carrying a passenger of mass 100 kg. It is moving upwards but decelerating (slowing down) at $2.5 \, \text{m/s}^2$. We are tasked with calculating the tension in the cable supporting the lift.

Step-by-step breakdown:

1. Total mass: $m_{\text{total}} = m_{\text{lift}} + m_{\text{passenger}} = 600 \, \text{kg} + 100 \, \text{kg} = 700 \, \text{kg}$

2. Net force due to deceleration: When the lift is moving upward but decelerating, the tension in the cable is affected by the deceleration. Using Newton's second law: $F_{\text{net}} = m_{\text{total}} \cdot a$ where $a = -2.5 \, \text{m/s}^2$ (negative because the lift is slowing down as it moves upward).

3. Gravitational force (weight): The total weight acting on the lift is: $F_{\text{gravity}} = m_{\text{total}} \cdot g = 700 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 6860 \, \text{N}$

4. Net force: The net force on the lift is due to both the gravitational force and the deceleration. Since the lift is decelerating upwards, the effective force that the tension needs to overcome is: $F_{\text{net}} = m_{\text{total}} \cdot (-a) = 700 \, \text{kg} \times (-2.5 \, \text{m/s}^2) = -1750 \, \text{N}$

5. Tension in the cable (T): The tension must balance the force of gravity and provide enough force to decelerate the lift: $T = F_{\text{gravity}} + |F_{\text{net}}|$ $T = 6860 \, \text{N} + 1750 \, \text{N} = 8610 \, \text{N}$

Thus, the tension in the cable is 8610 N, which corresponds to the second option.

Would you like further details or clarifications on this solution?

Here are 5 related questions:

1. How would the tension change if the lift was accelerating upwards at $2.5 \, \text{m/s}^2$?
2. How do you calculate the net force if the lift is in free fall?
3. What happens to the tension in the cable when the lift moves downward with constant velocity?
4. How does increasing the lift's mass affect the tension in the cable?
5. What would the tension be if the lift was moving upward at constant velocity?

Tip: Always use a free body diagram to visualize forces acting on objects when solving tension or acceleration problems!