Constant Resistance in Physics Problem

[p.mather]

Homework Statement

A metal stake of mass 8 kg is driven vertically into the ground using a hammer of 10 kg. Just prior to impact the speed of the hammer is 12 m/s. After impact the hammer remains in contact with the stake.

a. What is the speed that the stake starts to enter the ground?

b. During penetration the ground gives a constant resistance of 1500 N. How far does the stake penetrate the ground?

Homework Equations

I am having a problem working the second part of this question out. Please could somebody help. Thanks in advance.

The Attempt at a Solution

Part a)

Momentum before= 10kg x 12m/s + 8kg x 0m/s
Momentum after= 18kg x 'v'
So velocity=6.67m/s

Part b)

Help if possible.

 

[Disconnected]

This post was wrong.

 

[gneill]

I think it is assumed that all of the momentum is transferred to the stake and that the hammer no longer does any work on it.

From the problem statement:
"After impact the hammer remains in contact with the stake."

So the hammer and stake behave as a single object throughout the motion that occurs following impact.

Note that both the stake and hammer have weight...

 

[p.mather]

I understand that the acceleration is negative as you have said however...

Could somebody please explain the answer then as i am led to believe that the answer is 0.302m, however i am having trouble computing a answer that is even close to the answer.

Thanks for your help.

 

[gneill]

You have already calculated the initial velocity of the hammer+stake immediately following the impact using conservation of momentum.

What if you were to think in terms of conservation of energy from this point? The hammer+stake have an initial kinetic energy. Where will that energy go as the stake is driven in? Are there any other reservoirs of energy to consider?

 

[Disconnected]

I think it is assumed that all of the momentum is transferred to the stake and that the hammer no longer does any work on it.

From the problem statement:
"After impact the hammer remains in contact with the stake."

So the hammer and stake behave as a single object throughout the motion that occurs following impact.

Note that both the stake and hammer have weight...

Thanks for the catch, I got a little lazy minded there. Not to mention that I wrote COMPLETELY THE WRONG EQUATION. It should, of course, have been v^2=u^2+2as

So v=0, u=6.67, a=-1500/18
s=(6.67^2)/(2*1500/18)

 

[p.mather]

Im not 100% sure what you mean, as this area is not my strong point. Could you please explain.

 

[p.mather]

Thanks for the catch, I got a little lazy minded there. Not to mention that I wrote COMPLETELY THE WRONG EQUATION. It should, of course, have been v^2=u^2+2as

So v=0, u=6.67, a=-1500/18
s=(6.67^2)/(2*1500/18)

I thought you had the wrong equation. No worries. Thanks so working that equation out...

I seem to get a depth of 0.267m however i am led to believe that the answer is 0.302m... thoughts?

 

[Disconnected]

From what source are you getting the .302m?

 

[p.mather]

Its the answer that we have been given for the depth from a lecturer. However i can't seem to find that answer.

 

[gneill]

What is the initial kinetic energy of the hammer+stake ensemble?

 

[p.mather]

What is the initial kinetic energy of the hammer+stake ensemble?

Just out of interest do you find the answer to be 0.302m?

I have computed the kinetic energy of the hammer and stake but what's the reson for that?!

 

[Disconnected]

What is the initial kinetic energy of the hammer+stake ensemble?

I'm getting the same result using conservation of energy (ignoring g, I mean, surely not, right?).

 

[Disconnected]

Just out of interest do you find the answer to be 0.302m?

I have computed the kinetic energy of the hammer and stake but what's the reson for that?!

Conservation of energy.
The work done is equal to the force times the distance over which that force acts, and to the change in kinetic energy. In this case the change in kinetic energy is between the initial K energy and zero, and is thus just the initial K energy.
So

work=f*d
d=work/1500

 

[p.mather]

Conservation of energy.
The work done is equal to the force times the distance over which that force acts, and to the change in kinetic energy. In this case the change in kinetic energy is between the initial K energy and zero, and is thus just the initial K energy.
So

work=f*d
d=work/1500

Yes i see where you are coming from with that, as u say the answer still comes out to be 0.267m.

Maybe the 0.302m could be wrong however i would be suprised if it was given its come from the lecturer.

 

[gneill]

If the distance that the stake penetrates is d, then the stake+hammer both descend by that amount. The work done by friction is indeed f*d. But the lowering of the hammer+stake in the Earth's gravity field will supply additional energy to the initial store of kinetic energy. How much additional energy is available if the hammer+stake drop by in height distance d?

 

[p.mather]

If the distance that the stake penetrates is d, then the stake+hammer both descend by that amount. The work done by friction is indeed f*d. But the lowering of the hammer+stake in the Earth's gravity field will supply additional energy to the initial store of kinetic energy. How much additional energy is available if the hammer+stake drop by in height distance d?

So by doing it this method can u get the answer of 0.302m, if so could you explain. Thanks.

 

[Disconnected]

So by doing it this method can u get the answer of 0.302m, if so could you explain. Thanks.

I think they want you to answer the question. How much energy is gained?

 

[p.mather]

I think they want you to answer the question. How much energy is gained?

How does that calculate to 0.302 though?

 

[gneill]

The idea is to balance the energy equation. There are three places where energy is involved in this question: Initial kinetic energy, gravitational potential energy, and work done by friction. Both the work done and the gravitational energy depend upon the distance that the stake penetrates. Can you write a suitable equation?

 

[Disconnected]

How does that calculate to 0.302 though?

Well, if there is more energy, it will go farther, agreed? Let's find out how much more more energy it gains from moving through the g field, maybe it will cause it to move another 0.04m?

 

[p.mather]

Well, if there is more energy, it will go farther, agreed? Let's find out how much more more energy it gains from moving through the g field, maybe it will cause it to move another 0.04m?

Yes i agree with that, so how do..
KE=0.5mv^2
PE=mgh
work=force x distance

relate in this case?

 

[gneill]

The falling object begins with a store of kinetic energy, KE. As it moves downward it picks up additional energy from gravitation, mgh. It also loses energy to friction (work done), f*h. The stake stops moving when the store of energy is empty.

 

[KaiserBrandon]

yes, putting gravity into consideration will allow you to get the answer you are looking for. What is the force of gravity on the hammer stake system and therefore what is the net force acting upon it?

 

[Disconnected]

Yes i see where you are coming from with that, as u say the answer still comes out to be 0.267m.

Maybe the 0.302m could be wrong however i would be suprised if it was given its come from the lecturer.

holy embarrasing brainfart, batman!

The TOTAL force acting on the stake and hammer will be the sum of the resistant force and the force due to gravity!

How embarrasing:shy: