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 cable will be (`g=10 ms^(-2)`)

Options

A. 8000 N
B. 6400 N
C. 11200 N
D. 9600 N

Explanation

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

### Step 1: Identify the given data
- Total mass of the elevator and its load, \( m = 800 \, \text{kg} \)
- Initial velocity of the elevator, \( u = 10 \, \text{m/s} \) (downward)
- Final velocity of the elevator, \( v = 0 \, \text{m/s} \) (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
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Substituting the values:
\[ 0 = (10)^2 + 2a(25) \]
This simplifies to:
\[ 0 = 100 + 50a \]
Rearranging gives:
\[ 50a = -100 \quad \Rightarrow \quad a = -2 \, \text{m/s}^2 \]
The negative sign indicates that the acceleration is acting in the opposite direction to the motion (deceleration).

### Step 3: Calculate the tension in the supporting cable
The tension \( T \) in the cable can be calculated using the formula:
\[ T = m(g + a) \]
Where:
- \( g \) is the acceleration due to gravity (10 m/s²)
- \( a \) is the deceleration (which we found to be -2 m/s²)

Substituting the values:
\[ T = 800 \, \text{kg} \times (10 \, \text{m/s}^2 + (-2 \, \text{m/s}^2)) \]
This simplifies to:
\[ T = 800 \, \text{kg} \times (10 - 2) \, \text{m/s}^2 = 800 \, \text{kg} \times 8 \, \text{m/s}^2 \]
Calculating this gives:
\[ T = 6400 \, \text{N} \]

### Final Answer
The tension in the supporting cable is \( 6400 \, \text{N} \).

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