An elevator and its load have a total mass of 800 kg. If the elevator, originally moving downwards at `10 ms^(-1)`, is brought to rest with constant deceleration in a distance of 25 m, the tension in the supporting canble will be (`g=10 ms^(-2)`)

Options

A. 8000 N
B. 1600 N
C. 9600 N
D. 6400 N

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the known values
- Mass of the elevator and load, \( m = 800 \, \text{kg} \)
- Initial velocity, \( u = 10 \, \text{m/s} \) (downwards)
- Final velocity, \( v = 0 \, \text{m/s} \) (at rest)
- Distance over which the elevator comes to rest, \( s = 25 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the deceleration (acceleration) using the third equation of motion
Using the equation:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ 0 = (10)^2 + 2a(25) \]
\[ 0 = 100 + 50a \]
Rearranging gives:
\[ 50a = -100 \implies a = -2 \, \text{m/s}^2 \]
This indicates a deceleration of \( 2 \, \text{m/s}^2 \).

### Step 3: Analyze the forces acting on the elevator
The forces acting on the elevator are:
- Weight of the elevator, \( W = mg = 800 \times 10 = 8000 \, \text{N} \) (downwards)
- Tension in the cable, \( T \) (upwards)

### Step 4: Apply Newton's second law
Since the elevator is decelerating upwards, we can write:
\[ T - mg = -ma \]
Substituting the known values:
\[ T - 8000 = -800 \times (-2) \]
\[ T - 8000 = 1600 \]
Rearranging gives:
\[ T = 8000 + 1600 = 9600 \, \text{N} \]

### Final Answer
The tension in the supporting cable is \( T = 9600 \, \text{N} \).

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