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:
[tex]E = \frac{mv^2}{2} = 240000[/tex]
[tex]v^2 = \frac{240000*2}{3000} = 160[/tex]
Taking gravity into consideration, the net force acting on the pile once it hits the ground is
[tex] -5*10^6 + 10*3000 = -4970000 N[/tex]
Therefore the net deceleration is:
[tex]a = F/m = -49700003000 = -1657 m/s^2[/tex]
We can find out how far it's driven into the ground with the following equation of motion:
[tex]v^2 - v_0^2 = 2as[/tex]
where v = 0 is the final velocity, which is 0 because it stops, [tex]v_0^2 = 160[/tex] is the initial velocity when it hits the ground, a is the deceleration, and s is the distance it travels into the ground
[tex] 0 - 160 = 2*(-1657)s[/tex]
[tex]s = \frac{-160}{2*(-1657)} = 0.0483m[/tex] or 48.3 mm
