Question

A pile 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?

Answer 1

The gravitational potential energy becomes work driving the pipe 

m g h = f d 

450 * g * 1.5 = f * .45

Answer 2

Answer:

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

$vf^{2}=vi^{2}+2*g*dy$

We get the vf

$vf=\sqrt{2*g*dy}$

$vf=\sqrt{2*9.8m/s^{2}*1.5m }$

vf = 4.42m / s

Then the kinetic energy is equal to the work done.

$KE=\frac{1}{2}*v^{2} *m$

KE=1/2*$5.42m/s^{2}$*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