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. 6400 N
C. 11200 N
D. 9600 N

Explanation

To solve the problem step by step, we will use the equations of motion and the concepts of forces acting on the elevator.

### Step 1: Identify the given data
- Total mass of the elevator (m) = 800 kg
- Initial velocity (u) = 10 m/s (downward)
- Final velocity (v) = 0 m/s (at rest)
- Distance (s) = 25 m
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the acceleration (a) using the equation of motion
We will use the equation of motion:
\[ v^2 = u^2 + 2as \]
Rearranging it to find acceleration (a):
\[ a = \frac{v^2 - u^2}{2s} \]

Substituting the known values:
\[ a = \frac{0^2 - (10)^2}{2 \times 25} \]
\[ a = \frac{-100}{50} \]
\[ a = -2 \, \text{m/s}^2 \]

The negative sign indicates that the acceleration is in the opposite direction to the initial velocity (upward).

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

Using Newton's second law, the net force acting on the elevator can be expressed as:
\[ T - W = ma \]
Substituting the values:
\[ T - 8000 = 800 \times (-2) \]
\[ T - 8000 = -1600 \]
\[ T = 8000 - 1600 \]
\[ T = 6400 \, \text{N} \]

### Step 4: Final calculation of tension
Since we need to consider the upward acceleration while the elevator is decelerating, we need to add the weight to the force due to acceleration:
\[ T = W + ma \]
\[ T = 8000 + 1600 \]
\[ T = 9600 \, \text{N} \]

### Conclusion
The tension in the supporting cable when the elevator is brought to rest is **9600 N**.

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