Can someone help explain the normal force as the average force in a pile driver?

[LuckyIam]

Hey everyone, this is a copy/paste from a another thread. I actually don't need help with finding an answer. Its a practice problem in the book so its all worked out. What I need is for someone to help explain to me why the normal force is the average...I'm having trouble understanding this part1. Homework Statement

In a pile driver , a steel hammerhead with mass 200kg is lifted 3m above the top of a vertical I beam being driven into the ground. the hammer is then dropes driving the Ibeam7.4cm farther into the ground. The vertical rails that guide the hammerhead exert a constant 60N friction force on the hammerhead. Use the work energy theorem to find a) speed of the hammerhead just as it hits the I beam andb) the average force the hammerhead exerts on the I-Beam Ignore the air effects of the air.

The Attempt at a Solution

I managed to understand part a. Part b is where I am lost.

The book shows the following work, and I need help understanding on of the steps.

Wtotal = (w-f-n)s23

Wtotal = (w-f-n)s23 = k3-k2

*** n is the same as g right? and we solve for n? Since when is n the average force?***

n = w-f - k3-k2/s23

=1960N - 60N - 0J - 5700J / 0.074m

=7900N

the part underlined I understand.
The part in bold is where I can't explain why.

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[haruspex]

Wtotal = (w-f-n)s23

Wtotal = (w-f-n)s23 = k3-k2

I need to know what all these variables represent.

Read more: https://www.physicsforums.com

That link does not go to a specific thread.

 

[BvU]

the part underlined I understand

Good.

Well, let me guess: 2 is when the hammer hits, 3 is when the 7.4 cm are done. s₂₃ is then the 0.74 m.

So what you have is some kind of energy equation:
E_{kin, 2}+E_{potential, 2} = Work_beam + Work_fric +E_{potential, 3}

with Work_beam the energy used to drive the I beam down
and Work_fric the energy lost in friction.

Work_beam is s23 * average force, hence the average force.

Your answer is a factor 10 off - or you made a typo.
Probably forgot the brackets in

1960N - 60N - ( 0J - 5700J ) / 0.074m

AND a zero in the result.

That makes it time-consuming to help you !
Have pity with the poor helpers who desperately try to understand what you type: read what you typed as if you were one of them.