An elevator cab of mass `m=500` kg is descending with speed `v_(i)=4.0m//s` when its supporting cable begins to slip, allowing it to fall with constant acceleration `vec(a)=vec(g)//5`.
What is the cab's kinetic energy at the end of the 12 m fall ?

Explanation

To find the kinetic energy of the elevator cab at the end of a 12-meter fall, we can use the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the elevator cab, \( m = 500 \, \text{kg} \)
- Initial speed, \( v_i = 4.0 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
...