A lift of mass m is supported by a cable that can with stand a force of 3mg. Find the shortest distance in which the lift can be stopped when it is descending with a speed of g/4.

Answer

To solve the problem, we will follow these steps:

### Step 1: Analyze the Forces Acting on the Lift
The lift of mass \( m \) is descending with a speed of \( \frac{g}{4} \). The forces acting on the lift are:
- The weight of the lift acting downwards: \( mg \)
- The tension in the cable acting upwards: \( T \)

The cable can withstand a maximum tension of \( 3mg \).

### Step 2: Apply Newton's Second Law
Since the lift is descending, we can apply Newton's second law:
\[ T - mg = -ma \]
Here, \( a \) is the upward acceleration (deceleration in the context of stopping the lift).

### Step 3: Determine Maximum Tension
Given that the maximum tension \( T \) can be \( 3mg \), we substitute this into our equation:
\[ 3mg - mg = -ma \]
This simplifies to:
\[ 2mg = -ma \]
Rearranging gives:
\[ a = -2g \]
This means the lift is decelerating at \( 2g \) upwards.

### Step 4: Use Kinematic Equation to Find Distance
We need to find the shortest distance \( s \) in which the lift can be stopped. We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 0 \) (final velocity when stopped)
- \( u = \frac{g}{4} \) (initial velocity)
- \( a = -2g \) (deceleration)

Substituting the values into the equation:
\[ 0 = \left(\frac{g}{4}\right)^2 + 2(-2g)s \]
This simplifies to:
\[ 0 = \frac{g^2}{16} - 4gs \]
Rearranging gives:
\[ 4gs = \frac{g^2}{16} \]
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ 4s = \frac{g}{16} \]
Thus:
\[ s = \frac{g}{64} \]

### Final Answer
The shortest distance in which the lift can be stopped is:
\[ s = \frac{g}{64} \]

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