An elevator weighs `4000kg` When the upward tension in the supporting cable is `48000 N` what is the upward acceleration? Starting from rest, how far does it rise in `3` seconds ?

Explanation

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

### Step 1: Calculate the Weight of the Elevator
The weight (W) of the elevator can be calculated using the formula:
\[ W = m \cdot g \]
Where:
- \( m = 4000 \, \text{kg} \) (mass of the elevator)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 4000 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 40000 \, \text{N} \]

### Step 2: Determine the Net Force Acting on the Elevator
The net force (F_net) acting on the elevator can be calculated using the formula:
\[ F_{\text{net}} = T - W \]
Where:
- \( T = 48000 \, \text{N} \) (tension in the cable)
- \( W = 40000 \, \text{N} \) (weight of the elevator)

Calculating the net force:
\[ F_{\text{net}} = 48000 \, \text{N} - 40000 \, \text{N} = 8000 \, \text{N} \]

### Step 3: Calculate the Upward Acceleration of the Elevator
Using Newton's second law, we can find the acceleration (a) using the formula:
\[ F_{\text{net}} = m \cdot a \]
Rearranging gives:
\[ a = \frac{F_{\text{net}}}{m} \]

Substituting the values:
\[ a = \frac{8000 \, \text{N}}{4000 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Step 4: Calculate the Distance the Elevator Rises in 3 Seconds
Using the kinematic equation for distance (s) when starting from rest:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( u = 0 \, \text{m/s} \) (initial velocity)
- \( a = 2 \, \text{m/s}^2 \) (acceleration)
- \( t = 3 \, \text{s} \)

Substituting the values:
\[ s = 0 + \frac{1}{2} \cdot 2 \, \text{m/s}^2 \cdot (3 \, \text{s})^2 \]
\[ s = \frac{1}{2} \cdot 2 \cdot 9 = 9 \, \text{m} \]

### Final Answers:
- The upward acceleration of the elevator is \( 2 \, \text{m/s}^2 \).
- The distance it rises in 3 seconds is \( 9 \, \text{m} \).

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