A lift a mass `1200 kg` is raised from rest by a cable with a tension `1350 kg f` . After same time the tension drops to `1000 kg f` and the lift comes to rest at a height of `25 m` above its intial point `(1 kg - f = 9.8 N)`
What is the greatest speed of lift?

Options

A. `9.8 ms^(-1)`
B. `7.5 ms^(-1)`
C. `5.92 ms^(-1)`
D. None of these

Explanation

To solve the problem step by step, we will analyze the motion of the lift in two phases: the acceleration phase and the deceleration phase.

### Step 1: Calculate the acceleration when the tension is 1350 kgf

1. **Convert the tension from kgf to Newtons**:
\[ T_1 = 1350 \, \text{kgf} \times 9.8 \, \text{N/kgf} = 13230 \, \text{N} \]

2. **Calculate the weight of the lift**:
\[ W = m \cdot g = 1200 \, \text{kg} \times 9.8 \, \text{N/kg} = 11760 \, \text{N} \]

3. **Determine the net force acting on the lift**:
\[ F_{\text{net}} = T_1 - W = 13230 \, \text{N} - 11760 \, \text{N} = 1470 \, \text{N} \]

4. **Calculate the acceleration using Newton's second law**:
\[ a_1 = \frac{F_{\text{net}}}{m} = \frac{1470 \, \text{N}}{1200 \, \text{kg}} = 1.225 \, \text{m/s}^2 \]

### Step 2: Calculate the distance traveled during acceleration

1. **Use the kinematic equation to find the distance (h1) during acceleration**:
Since the lift starts from rest, we can use the formula:
\[ h_1 = \frac{1}{2} a_1 t^2 \]
We will need to find the time (t) later.

### Step 3: Calculate the deceleration when the tension drops to 1000 kgf

1. **Convert the tension from kgf to Newtons**:
\[ T_2 = 1000 \, \text{kgf} \times 9.8 \, \text{N/kgf} = 9800 \, \text{N} \]

2. **Determine the net force during deceleration**:
\[ F_{\text{net}} = T_2 - W = 9800 \, \text{N} - 11760 \, \text{N} = -1960 \, \text{N} \]

3. **Calculate the deceleration**:
\[ a_2 = \frac{F_{\text{net}}}{m} = \frac{-1960 \, \text{N}}{1200 \, \text{kg}} = -1.6333 \, \text{m/s}^2 \]

### Step 4: Calculate the distance traveled during deceleration (h2)

1. **The total height the lift travels is 25 m**:
\[ h_1 + h_2 = 25 \, \text{m} \]

2. **Using the kinematic equation for deceleration**:
\[ v^2 = u^2 + 2a_2 h_2 \]
where \( u \) is the maximum speed reached at the end of the acceleration phase.

### Step 5: Relate the maximum speed to the distances and accelerations

1. **Using the relationship**:
\[ v^2 = 2a_1 h_1 = -2a_2 h_2 \]

2. **From the previous steps, we can express \( h_2 \) in terms of \( h_1 \)**:
\[ h_2 = 25 - h_1 \]

### Step 6: Solve for maximum speed (v)

1. **Substituting the values**:
\[ v^2 = 2 \cdot 1.225 \cdot h_1 = -2 \cdot (-1.6333) \cdot (25 - h_1) \]

2. **Solving the equations simultaneously will give us the maximum speed**:
After calculations, we find:
\[ v = 5.92 \, \text{m/s} \]

### Final Answer
The greatest speed of the lift is \( 5.92 \, \text{m/s} \).

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