A 1,900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.80 m before coming into contact with the top of the beam, and it drives the beam 14.2 cm farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

I was able to figure out that the magnitude is 516901 N and the direction is upward by using a similar question but I would like a step-by-step explanation on how I got this using:

mgh=average force x d

Question

A 1,900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.80 m before coming into contact with the top of the beam, and it drives the beam 14.2 cm farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

I was able to figure out that the magnitude is 516901 N and the direction is upward by using a similar question but I would like a step-by-step explanation on how I got this using: 

mgh=average force x d

anonymous · Answer

And why is it average force x d=mgh rather than PE=mgh

jfraser · Answer

when an object is moved against gravity, work is done, and it has units of energy. The work-energy theorem states that the change in PE is equal to the work done, so $$mg\Delta h = F*d$$ when the piledriver is falling, its PE changes by $mg\Delta h$, so it has that much KE when it strikes the beam. The force it applies will cause the beam to drop by the distance d, leaving the only missing info the force of collision

jfraser · Answer

the trick is that the $\Delta h$ of the piledriver and the d of the work done are not the same distances