# Homework Help: Energy, non-conservative and conservative forces question

1. Dec 31, 2009

### holezch

1. The problem statement, all variables and given/known data

An object with an initial velocity V0 of 14 m/2 falls from a height of 240m and buries itself in 0.20 m of sand. The mass of the body is 1.0 kg. Find the average resistive force exerted by the sand on the body. Neglect air resistance.

2. Relevant equations

change in K = W (of resultant force)
change in K + change in U (potential energy) = 0 [conservative forces]
change in K + change in U + work done by friction = 0 [non conservative friction]\

...et c

3. The attempt at a solution

Solution from textbook:

the kinetic energy of the body as it is about to hit the sand:

K = (mV0^2)/2 + mgh

where m is mass g is gravity h is the height from the sky to the sand

Also, from the work-energy principle: (then it says "approximately")

K = Fs, where F is the average resistive force of the sand onto the body and s is the distance into the sand.

This part is weird to me.. shouldn't change in K equal to the work done as the object falls from the sky? How come K = Fs?

Then they just plug things in and solve.. I just don't get how they could just ignore the non-conservative force from friction and how they could just say that K = Fs..

2. Dec 31, 2009

### PhanthomJay

It's a bit confusing the way the problem was solved. They first considered the KE of the object just before it hit the ground, using conservation of Mechanical Energy with only conservative forces acting (there is no friction during the fall, neglecting air resistance). That's your second equation. Then the second part uses the work enrgy theorem, and neglects the work done by gravity in that short distance through the sand (W_c +W_nc = delta KE, where W_c is neglected). The more exact solution utilizes your 3rd equation, from start point to end point.

3. Dec 31, 2009

### holezch

Hi, thank you so much for replying.

I think I'm still confused:

W = Fs = change in K, K1 - K0..
We have K0 as it's about to go into the ground, K0 = (mV0^2)/2 + mgh.

K1, we don't know, but it should be something like W (by conservative forces) + W (by friction) + K0.

But then... K0 = Fs.

Also, how can they just omit things? Maybe it's due to the fact that it's the average? I can't quite what the difference between finding the average and finding the precise force would be.. Is it just average because we're assuming that the sand resistance is always the same?

Thank you, this is as much as I understand right now :S thank you for your patience.

EDIT: Okay, so they some how neglected the gravity, and they got K1 - K0 = work by friction = Fs. Still, that would be K1 - K0, and not K0 = Fs? [I posted the question above denoting K0 as K]

Last edited: Dec 31, 2009
4. Dec 31, 2009

### PhanthomJay

Correct
you do know K1, it comes to a stop after burying itself 0.2 m into the sand, K1 = ???
No, K1 -K0 = Fs, as you noted above
Nothing is being omitted, except they are neglecting the work done by gravity in the short 0.2 m distance, which is small in comparision to the work done by the resistive sand force.
It's average because the sand resistance is not constant, it is variable, from 0 when the object first hits, to a maximum at 0.2 m (just like a spring force obeying Hooke's law, F = kx)

5. Dec 31, 2009

### holezch

ahh , okay. So K1 = 0?

Then 0 - K0 = -((mV0^2)/2 + mgh) = Fs?

The book says (mV0^2)/2 + mgh) = Fs..

Okay, I understand everything else :) Thanks so much!

6. Dec 31, 2009

### PhanthomJay

You are correct, the work done by the sand resistive force is negative (the force acts up, the displacement is down, so the work done by the sand resistive force is negative). The magnitude of the force is a positive number.

7. Dec 31, 2009

### holezch

Excellent!

Thank you so much :) and I especially appreciate you helping me during new year's eve :)

Happy new year!

8. Dec 31, 2009

### PhanthomJay

Same to you. I hope I can stay awake to watch the ball come down in Times Square, but only to be disappointed again for the umpteenth time. I really want to see Dick Clark, though...grew up with the guy watching him on American Bandstand. Have a Happy!!

Edit: And what a disappointment it was, they didn't even show the ball at midnight, its all computerized at a cost of millions, and I don't think the thing even drops anymore...Bring back the old days when the ball cost 10 bucks and did something, and American Bandstand and Dick Clark were in their prime.....

Last edited: Jan 1, 2010