To solve the problem of finding the shortest distance in which a lift can be stopped when descending with a speed of 2.5 m/s, we can follow these steps:
### Step 1: Identify the Forces Acting on the Lift
The lift has a mass \( m = 1000 \, \text{kg} \). The gravitational force acting on the lift is given by:
\[
F_g = mg = 1000 \, \text{kg} \times 10 \, \text{m/s}^2 = 10000 \, \text{N}
\]
The cable can support a maximum load of \( 2000 \, \text{kg} \), which translates to a maximum tension \( T \) in the cable:
\[
T_{\text{max}} = 2000 \, \text{kg} \times 10 \, \text{m/s}^2 = 20000 \, \text{N}
\]
### Step 2: Write the Equation of Motion
When the lift is descending and needs to stop, the tension \( T \) in the cable acts upwards while the gravitational force acts downwards. The net force can be expressed as:
\[
T - mg = ma
\]
Where \( a \) is the upward acceleration (or deceleration in this case). Rearranging gives:
\[
T = mg + ma
\]
### Step 3: Substitute Maximum Tension
To find the maximum deceleration, we can set \( T \) to its maximum value:
\[
20000 \, \text{N} = 10000 \, \text{N} + 1000 \, \text{kg} \cdot a
\]
Solving for \( a \):
\[
20000 \, \text{N} - 10000 \, \text{N} = 1000 \, \text{kg} \cdot a
\]
\[
10000 \, \text{N} = 1000 \, \text{kg} \cdot a
\]
\[
a = \frac{10000 \, \text{N}}{1000 \, \text{kg}} = 10 \, \text{m/s}^2
\]
### Step 4: Use Kinematic Equation to Find Stopping Distance
Now we can use the kinematic equation to find the stopping distance \( S \):
\[
v^2 = u^2 + 2aS
\]
Where:
- \( v = 0 \, \text{m/s} \) (final velocity when the lift stops)
- \( u = -2.5 \, \text{m/s} \) (initial velocity, negative because it's downward)
- \( a = 10 \, \text{m/s}^2 \) (deceleration is positive in the upward direction)
Substituting these values into the equation:
\[
0 = (-2.5)^2 + 2 \cdot 10 \cdot S
\]
\[
0 = 6.25 + 20S
\]
\[
20S = -6.25
\]
\[
S = -\frac{6.25}{20} = -0.3125 \, \text{m}
\]
### Step 5: Interpret the Result
The negative sign indicates that the distance is in the direction opposite to the initial motion (upward). Thus, the shortest distance in which the lift can be stopped is:
\[
S = 0.3125 \, \text{m}
\]
### Final Answer
The shortest distance in which the lift can be stopped is \( 0.3125 \, \text{m} \).
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