To solve the problem step by step, we will analyze the motion of the lift under two different conditions of tension.
### Step 1: Identify the forces acting on the lift during the first phase of motion.
The lift has a mass of \( m = 1200 \, \text{kg} \). The tension in the cable when the lift is being raised is \( T_1 = 1350 \, \text{kgf} \). We need to convert this tension into Newtons:
\[
T_1 = 1350 \, \text{kgf} \times 9.8 \, \text{N/kg} = 13230 \, \text{N}
\]
The weight of the lift is:
\[
W = m \cdot g = 1200 \, \text{kg} \times 9.8 \, \text{N/kg} = 11760 \, \text{N}
\]
### Step 2: Calculate the net force and acceleration during the first phase.
The net force \( F_{net1} \) acting on the lift when it is being raised is:
\[
F_{net1} = T_1 - W = 13230 \, \text{N} - 11760 \, \text{N} = 1470 \, \text{N}
\]
Using Newton's second law, we can find the acceleration \( a_1 \):
\[
F_{net1} = m \cdot a_1 \implies a_1 = \frac{F_{net1}}{m} = \frac{1470 \, \text{N}}{1200 \, \text{kg}} = 1.225 \, \text{m/s}^2
\]
### Step 3: Analyze the second phase of motion when the tension drops.
When the tension drops to \( T_2 = 1000 \, \text{kgf} \):
\[
T_2 = 1000 \, \text{kgf} \times 9.8 \, \text{N/kg} = 9800 \, \text{N}
\]
Now, we calculate the net force \( F_{net2} \) when the lift is coming to rest:
\[
F_{net2} = T_2 - W = 9800 \, \text{N} - 11760 \, \text{N} = -1960 \, \text{N}
\]
This indicates that the lift is decelerating. The acceleration (or retardation) \( a_2 \) can be calculated as:
\[
F_{net2} = m \cdot a_2 \implies a_2 = \frac{F_{net2}}{m} = \frac{-1960 \, \text{N}}{1200 \, \text{kg}} = -1.6333 \, \text{m/s}^2
\]
### Step 4: Use kinematic equations to relate the heights and accelerations.
Let \( h_1 \) be the height covered during the first phase and \( h_2 \) be the height covered during the second phase. The total height is given as:
\[
h_1 + h_2 = 25 \, \text{m}
\]
Using the kinematic equations, we have:
\[
v^2 = 2a_1 h_1 \quad \text{(1)}
\]
\[
0 = v^2 - 2a_2 h_2 \quad \text{(2)}
\]
From equation (1), we can express \( v^2 \):
\[
v^2 = 2a_1 h_1
\]
From equation (2), we can express \( v^2 \) as well:
\[
v^2 = 2a_2 h_2
\]
Setting these two expressions for \( v^2 \) equal gives:
\[
2a_1 h_1 = 2a_2 h_2
\]
\[
\frac{h_1}{h_2} = \frac{a_2}{a_1}
\]
### Step 5: Substitute values and solve for \( h_1 \) and \( h_2 \).
Substituting the values of \( a_1 \) and \( a_2 \):
\[
\frac{h_1}{h_2} = \frac{1.6333}{1.225} \approx 1.3333
\]
Let \( h_1 = 1.3333 h_2 \). Now substituting into the height equation:
\[
1.3333 h_2 + h_2 = 25
\]
\[
2.3333 h_2 = 25 \implies h_2 \approx 10.71 \, \text{m}
\]
\[
h_1 = 25 - h_2 \approx 25 - 10.71 \approx 14.29 \, \text{m}
\]
### Final Answer:
The height at which the tension changes is approximately \( \boxed{14.3 \, \text{m}} \).
