Answer:
0.0483m or 48.3 mm
Explanation:
Let g = 10m/s2
At 8m, the pile has a potential energy of
P = mgh = 3000*10*8 = 240000 J
This energy is converted to kinetic energy once the pile drops down to the bottom:
![E = \frac{mv^2}{2} = 240000](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bmv%5E2%7D%7B2%7D%20%3D%20240000)
![v^2 = \frac{240000*2}{3000} = 160](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cfrac%7B240000%2A2%7D%7B3000%7D%20%3D%20160)
Taking gravity into consideration, the net force acting on the pile once it hits the ground is
![-5*10^6 + 10*3000 = -4970000 N](https://tex.z-dn.net/?f=%20-5%2A10%5E6%20%2B%2010%2A3000%20%3D%20-4970000%20N)
Therefore the net deceleration is:
![a = F/m = -49700003000 = -1657 m/s^2](https://tex.z-dn.net/?f=a%20%3D%20F%2Fm%20%3D%20-49700003000%20%3D%20-1657%20m%2Fs%5E2)
We can find out how far it's driven into the ground with the following equation of motion:
![v^2 - v_0^2 = 2as](https://tex.z-dn.net/?f=v%5E2%20-%20v_0%5E2%20%3D%202as)
where v = 0 is the final velocity, which is 0 because it stops,
is the initial velocity when it hits the ground, a is the deceleration, and s is the distance it travels into the ground
![0 - 160 = 2*(-1657)s](https://tex.z-dn.net/?f=%200%20-%20160%20%3D%202%2A%28-1657%29s)
or 48.3 mm