Question

What is the kinetic energy of an object with a mass of `1 kg` that has been in freefall for `2 s`?

Answer 1

I found `192J`

Explanation:

We can either:

1) use kinematic to find the final velocity after `2s`:
using: `v_f=v_i+at`
where `v_i` is the initial velocity (assumed as zero);
`a` is the acceleration of gravity (directed downward).
We get:
`v_f=0-9.8*2=19.6m/s`
From this we get:
Kinetic Energy: `K=1/2mv^2=1/2*1*(19.6)^2=192J`

or:

2) use kinematic (and Conservation of Energy) to find the vertical distance described after `2s` and the Potential Energy:
using: `y_f-y_i=v_it+1/2at^2`
Where we set `y_f=0` as the ground (zero Potential Energy).
giving:
`0-y_i=0*2-1/2*9.8*(2)^2`
so that the height (from ground) will be:
`y_i=19.6m`
At this position the object had a Potential Energy of:
`U=mgy_i=1*9.8*19.6=192J` that will be converted entirely into Kinetic Energy during the fall!