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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?

Verified Answer :

syed514

the gravitational potential energy becomes work driving the pipe 

m g h = f d 

450 g 1.5 = f * .45

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