How Far Does the Pile Move with Each Blow of the Hammer?

[BEANTOWN]

Homework Statement

A pile driver has a hammer head of mass 4000 kg which is repeatedly raised 3.0 m above a pile and dropped onto it. The force required to drive the pile downwards into the ground is 500 000 N. When the hammerhead is dropped, all the energy foes into moving the pile. How far does the pile go with each blow?

Homework Equations

W= Fd
Et= Eg + Ek
Et= mgh + .5mv^2

The Attempt at a Solution

I found the total energy when the hammer is fully raised.

Et= mgh + .5mv^2
Et= (4000)(9.8)(3) + .5(4000((0)?2
Et= 117 600 J

then i did d= W divided by force to get distance

by slash isn't working for some reason so
117600 divided by 500 000=
=.2352 m
= 23.52 cm

The answer is supposed to be 25.5 cm though. I don't know what to do...

 

[PhanthomJay]

What about the hammer weight? It has potential energy at the start of the pile motion.

 

[BEANTOWN]

I included it in the calculation. Please show me what you mean.

 

[PhanthomJay]

Your initial calculation had h = 3 m. Using PE = mgh where h =3, gives the PE above the pile top as the reference point. That's fine for calculating the total energy just before the hammer hits the pile.. But when it hits it, the pile moves to a depth 'x', and thus the PE change must be referenced to the depth x below the pile top. So using the instant the hammer hits the pile as the initial point, its energy is 117 600 plus mgx, and the work done on the pile from the ground is 500 000(x).

 

[asteriatic]

Your approach is correct, but there may be a slight rounding error in your calculations. I would recommend double checking your calculations and using more precise values for the constants (such as 9.81 for the acceleration due to gravity). Alternatively, you could also use the equation Et=Fd to find the distance, where F is the force and d is the distance. In this case, the distance would be:

d= Et/F
d= 117600/500000
d= 0.2352 m
d= 23.52 cm

This matches your calculation, so it is likely just a matter of rounding. However, if you want to get a more precise answer, you can use more precise values for the constants or use the equation Et=Fd.