Temp rise due to hammering nails

[Bradracer18]

Here's my question...

The 1.20 kg head of a hammer has a speed of 6.5m/s just before it strikes a nail and is brought to rest. Estimate the temperature rise of a 14 gram iron nail generated by ten such hammer blows done in quick succession. Assume the nail absorbs all the energy.


The picture shows a hammer hitting a nail with the swings being horizontal(as if you were nailing into a wall...not the floor).


So, I can't seem to get my formulas set to get past a certain point. This is what I've done so far(which is not much).

W= 4.186 J = 1 cal of heat
W = Fd
F=ma------where I get stuck...I've got velocity, but not acceleration.

Thanks for any help...I'd be glad to work it with some help...or maybe I can work it a little further until I get stuck again(if I do).

Brad

 

[Doc Al]

How much kinetic energy does the hammer possesses before it hits the nail? What happens to that energy?

 

[Bradracer18]

Ok...well I kinda see where you are going, but still need a bit more direction.

KE=1/2mv^2

KE(hammer)=25.35 J -----So I think this energy is now transferred into the nail, with some lost to heat?? Or, do I have to recaculate the KE for the nail(using its mass)?


I'm also not real sure how this energy(J) is going end up giving me a temp measurment...will I eventually relate it to the specific heat of iron?

Thanks again and I'll keep working...

Brad

 

[Doc Al]

So I think this energy is now transferred into the nail, with some lost to heat?? Or, do I have to recaculate the KE for the nail(using its mass)?

What's the speed of the nail after you hit it with a hammer? (Hint: Nothing to calculate.)

I'm also not real sure how this energy(J) is going end up giving me a temp measurment...will I eventually relate it to the specific heat of iron?

Absolutely.

 

[Bradracer18]

The speed of the hammer would be directly transferred to the nail...so it would be 6.5m/s. With that in mind(and the nails mass), I found KE(nail)=.296 J

Do I multiply 10 swings into both the KE of the nail and the KE of the hammer?? I'm thinking I do, it just seems obvious. Then I'm thinking I have to Joule measurements...so then do I multiply them both by the specific heat of iron...and add them together to obtain a final amount of heat created?

Thanks again!

 

[Doc Al]

The speed of the hammer would be directly transferred to the nail...so it would be 6.5m/s.

No! You bang a nail into the wall, does it goes shooting through the wall? (Let's hope not.) No, when the hammer hits, the nail moves a bit, then hammer and nail both stop. The KE of the nail is zero. (That's why I said there is nothing to calculate. )

All of the KE of the hammer gets absorbed by the nail as thermal energy.

 

[Bradracer18]

Ok...that does make sense, as I think of it more and more.

So, now I am thinking that I multiply my KE(hammer) by 10 swings to find total KE. Then do I use an equation similar to U(internal energy) = N(1/2mv^2)...where N is number of molecules?? This doesn't seem like the right equation though. Also, I can't yet understand why they gave me the mass of the nail(and maybe that is for later in the problem).

 

[Delzac]

[tex] \frac{1}{2} m v^2 = mc \theta[/tex]

 

[Doc Al]

So, now I am thinking that I multiply my KE(hammer) by 10 swings to find total KE.

Right.

Then do I use an equation similar to U(internal energy) = N(1/2mv^2)...where N is number of molecules?? This doesn't seem like the right equation though.

It's not. Think specific heat of iron, which you mentioned earlier. (You'll need the mass of the nail.)

 

[Bradracer18]

Ok, well I am thinking this now.

KE(total) = 253.5 J

Q = mc*(change in T)

Change in T = Q/mc

253.5 J * .014 Kg * 450 J/Kg*deg C = 40.24 deg C

Even close?

 

[Bradracer18]

yeah I knew they did heat up...as I used to frame houses/apartment buildings for a few yrs...

 

[Bradracer18]

nevermind...ha...just noticed the problem was odd, so I checked it in the back, and I have the correct answer!

Thanks a ton Doc Al!

Brad