What Determines the Force Exerted by a Hammerhead in a Pile Driver?

[Heat]

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 and b) 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.

 

[mgb_phys]

Consider the force exerted by the pile on the hammer.
You are slowing the hammer from the speed found in part A to 0 in 7.4cm.
Just work out the accelearation ( v^2=u^2+2as) and use f=ma to get the force, assuming the deceleration is uniform.

 

[ogb p]

I would like to clarify a few things about the solution provided in the book. First, the use of "n" in the work equation is not the same as the gravitational constant "g". In this context, "n" represents the normal force exerted by the I-beam on the hammerhead, which is equal to the weight of the hammerhead (mg) minus the friction force (f).

Secondly, the work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. In this case, the net work done on the hammerhead is equal to the work done by the weight (mg) minus the work done by friction (f) and the work done by the normal force (n). Therefore, we can write the equation as:

Wtotal = (mg - f - n)s23 = K3 - K2

Where K3 is the final kinetic energy of the hammerhead just as it hits the I-beam and K2 is its initial kinetic energy (which is zero).

To find the average force exerted by the hammerhead on the I-beam, we can rearrange the equation as:

n = mg - f - (K3 - K2)/s23

= (200 kg)(9.8 m/s^2) - 60 N - (0 J - 5700 J)/(0.074 m)

= 1960 N - 60 N - 77027.03 N

= 7900 N

So, the average force exerted by the hammerhead on the I-beam is 7900 N. This is the force that is required to drive the I-beam further into the ground.

I hope this explanation helps to clarify the solution provided in the book. If you have any further questions, please feel free to ask. I am here to help you understand the concepts and principles involved in this problem.