a pile drive has a 413 kg block that is lifted 1.4m and then dropped onto the pile being driven into the ground. suppose that it takes 24ms to stop when it hits the pile. what is the average force exerted on the pile by the pile driver?

Answer 1

Force x distance driven =
Potential energy of the lifted weight
= (413)(9.8)(1.4) J

The distance X that the pile gets driven each time is (1/2)*(pile impact velocity)*time

The impact velocity of the pike is
V = sqrt(2gH)

Compute V, X, and then solve for the force.

Answer 2

I just realized that this problem can be solved more directly using momentum considerations. The answer should be the same.
(Force)(stopping time)
= momentum at impact = M V

F = M V/(0.024 s)

V = sqrt(2gH), as before

Answer 3

A spelunker (cave explorer) drops a stone from rest into a hole. The speed of sound is 343 m/s in air, and the sound of the stone striking the bottom is heard 3.87 s after the stone is dropped. How deep is the hole?