GREPhysics.NET: GR8677 #8

GR8677 #8

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Mechanics $\Rightarrow$ }Conservation

Straight-forward conservation of energy problem:

$E_0=\frac{1}{2}m v^2 = E_f=Fl, <br />$

where $l=0.025 m =\frac{25}{1000}$ , $m=5kg$ , and $v=10 m/s$ . Plug everything in, and solve for $F$ , to get choice D, $10,000 N$ .

Alternate Solutions

dcn 2008-03-26 09:21:36	Correct me if I'm wrong but isn't Average Force defined simply as delta p/delta t. Since the stone comes to rest, assuming all momentum is transfered to the nail, the momentum of the stone just before it hits the nail is delta p. For delta t we can use the average v which is (10-0)/2 = 5m/s. Then delta t = y/(average v) =0.025/5. Plugging this in: Average F = (5kg)(10m/s)(5m/s)/(25x10^-3m)=10^4N. Reply to this comment
jcain6 2005-12-09 05:12:15	you can also use kinematics here. Use the velocity/position kinematic to solve for a: a = 2000 m/s^2 where you use 0.025m as change in y. F = ma = (5kg)(200m/s^2)= 10000N Reply to this comment

Comments

dcn
2008-03-26 09:21:36

Correct me if I'm wrong but isn't Average Force defined simply as delta p/delta t. Since the stone comes to rest, assuming all momentum is transfered to the nail, the momentum of the stone just before it hits the nail is delta p. For delta t we can use the average v which is (10-0)/2 = 5m/s. Then
delta t = y/(average v) =0.025/5.
Plugging this in:
Average F = (5kg)(10m/s)(5m/s)/(25x10^-3m)=10^4N.

Imperate
2008-09-11 12:23:00

This is the way I did it.rnIf the block has momentum $p=10*5=50kgms^{-1}$ Then this momentum must be reduced to zero over the course of the nail being driven in. The time it the nail travels can be found from $\Delta t=\frac{0.025m}{v_{av}}=\frac{0.025m}{5ms^{-1}}=\frac{25}{1000*5}=0.005s.$ Which implies $F=\frac{\Delta P}{\Delta t} =\frac{50}{0.005}=10,000N$ rnrnSo it can be done this way, but I wish I thought of using energy at the time.

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Richard
2007-10-31 12:36:30

I think it would be better phrased as a Work Problem. It's not really energy conservation, because energy is clearly not conserved.
Anyway, not that it makes any difference to the solution really, but you forgot to include the potential energy term $Mgh$ . It only contributes 50 J or so...so who cares right?

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kicksp
2007-10-29 09:33:05

I have to disagree with everyone. Kinematics give the average net force as 100000/9.81 N $\approx$ 10190 N (downward), but the question asks for the average net force exerted BY the stone on the nail. (The binomial theorem will be useful to make this approximation). The average frictional force acting on the nail is $\Delta E/d$ = 10000N (upward) from the expression for the mechanical energy lost due to a constant dissipative force. The magnitude of the force exerted on the nail BY the stone is therefore approximately 190 N. As an order of magnitude answer, then, the best anwer is B.

Go maroons!

kicksp
2007-10-29 13:05:19

Sorry, I read the problem wrong. the average force exerted ON the stone BY the nail is 10000 N:

$F = \frac{mv^2}{2\Delta x}$

By Newton's Third Law, this is the magnitude of the average force exerted ON the nail BY the stone. Still, the analysis by energy conservation is erroneous.

kicksp
2007-10-29 13:07:18

Using the work-energy theorem is more appropriate.

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kicksp
2007-10-29 09:32:30

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KarstenChu
2007-04-01 00:00:39

Respectfully, I must argue these solutions.

First of all, I do not believe we can apply conservation of energy to this problem. It's possible that its just really late and I am really tired, but it seems to me that energy isn't necessarily conserved here since friction is slowing down the nail and dissipating energy in heat and sound. Additionally, the problem does not say that the collision between the stone and the nail is elastic so I see no reason to attack this problem from a conservation of energy foundation.

I do believe that conservation of momentum, though, is the way to go. I used two equations in my solution, x=v0t-1/2at^2 and J = Ft = delta p = m delta v.

Given that x = .025, delta v = 10 = v0 and F=ma, I solved for t in the impulse equation and plugged into the kinematic eqn to solve for F.

hungrychemist
2007-08-26 19:18:40

This question cannot be solved exactly. Your kinematic approach is only approximation since one cannot assume the constant acceleration of the pin. Like Yosun said, energy conservation approach is the way to go although energy is not really conserved here you can see from the choices, within a reasonable loss of energy due to (friction and sound and so on) you can conclude the ans to be D.

TigerTed8
2007-09-17 19:36:41

Karsten, I did the problem the same way as you when I first approached it. However, I'm now a convert to the energy method. The friction of the nail, clearly non-trivial, is not a part of the equation here. The stone has energy, and then it loses that energy. This occurs primarily through some force. So $F \Delta x$ = $U_{stone}$ . The only way that the stone could lose energy that wasn't related to a force would be IR radiation, or the sound or heat from the stone on nail collision. The nail-wood friction occurs AFTER the transfer of energy from stone to nail by way of $F$ . Hope this helps.

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jcain6
2005-12-09 05:13:08

typo - below should read

F = ma = (5kg)(2000 m/s^2) = 10000

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jcain6
2005-12-09 05:12:15

you can also use kinematics here.

Use the velocity/position kinematic to solve for a:

a = 2000 m/s^2

where you use 0.025m as change in y.

F = ma = (5kg)(200m/s^2)= 10000N

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Ben
2005-12-02 11:19:26

I think m was mass not meter, otherwise it made no sense at all.

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Ben
2005-12-01 13:11:39

Should it not be 100m/2? As written, the answer would be 1,000 N...

daschaich
2005-12-01 21:45:21

Actually, it should be $10 m/s$ (dimensions and all that). Just a simple typo. Note that v is the initial speed and not the average speed. With $v=10$ ( $v^2=100$ ), the equation becomes $F=\frac{5*100*1000}{25*2}=10,000 N$ .

yosun
2005-12-02 01:10:22

ben, daschaich: thanks for the typo-alert; it's been fixed now. (note to self: always ignore dimensions when doing the numerical calculations.)

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