A 1000 kg lift is supported by a cable that can support 2000 kg. The shortest distance in which the lift can be stopped when it is descending with a speed of `2.5 m//s` is `( g = 10 m//s^(2))`

Options

A. `( 5)/( 16) m`
B. `( 5)/( 32) m`
C. `1m `
D. `2m`

Answer

To solve the problem, we need to determine the shortest distance in which a 1000 kg lift can be stopped when it is descending with a speed of 2.5 m/s. We will use the concepts of forces, acceleration, and equations of motion.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Lift:**
- The weight of the lift (W) is given by:
\[ W = mg = 1000 \, \text{kg} \times 10 \, \text{m/s}^2 = 10000 \, \text{N} \]
- The maximum tension (T) the cable can support is:
\[ T = 2000 \, \text{kg} \times 10 \, \text{m/s}^2 = 20000 \, \text{N} \]

2. **Determine the Net Force Acting on the Lift:**
- When the lift is descending and we want to stop it, the tension in the cable must overcome its weight. The net force (F_net) acting on the lift can be expressed as:
\[ F_{\text{net}} = T - W = 20000 \, \text{N} - 10000 \, \text{N} = 10000 \, \text{N} \]

3. **Calculate the Acceleration of the Lift:**
- Using Newton's second law, \( F = ma \), we can find the acceleration (a):
\[ F_{\text{net}} = ma \implies 10000 \, \text{N} = 1000 \, \text{kg} \times a \]
\[ a = \frac{10000 \, \text{N}}{1000 \, \text{kg}} = 10 \, \text{m/s}^2 \]

4. **Use the Equation of Motion to Find the Stopping Distance:**
- We will use the equation:
\[ v^2 = u^2 + 2as \]
- Here, \( v = 0 \, \text{m/s} \) (final velocity when stopped), \( u = 2.5 \, \text{m/s} \) (initial velocity), and \( a = -10 \, \text{m/s}^2 \) (deceleration, negative because it is stopping).
- Rearranging the equation gives:
\[ 0 = (2.5)^2 + 2(-10)s \]
\[ 0 = 6.25 - 20s \]
\[ 20s = 6.25 \implies s = \frac{6.25}{20} = 0.3125 \, \text{m} \]

5. **Convert to Fraction:**
- The distance can also be expressed as:
\[ s = \frac{5}{16} \, \text{m} \]

### Final Answer:
The shortest distance in which the lift can be stopped is \( \frac{5}{16} \, \text{m} \).

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