Apile driver lifts a 450 kg weight and then lets it fall onto the end of a steel pipe that needs to be driven into the ground. a fall of 1.5 m drives the pipe in 45 cm.
what is the average force exerted on the pipe?

F = 14700N

Explanation:

Hello! Let's solve this!

We have the following data:

vi = 0m / s

dy = 1.5m

m = 450kg

dx = 0.45m

First we need to know the speed of the impeller.

We can use the following formula

We get the vf

vf = 4.42m / s

Then the kinetic energy is equal to the work done.

KE=1/2**450kg

We cleared KE

KE = W = 6615J

From the working formula we will clear the force exerted.

W = F * dx

F = W / dx = 6615J / 0.45m

F = 14700N

i don't understand. explain it clearly