# Forces of a Hammer Acting On a Nail With constant acceleration

1. Feb 5, 2012

### LoveHOPEpeace

1. The problem statement, all variables and given/known data
A 4.9-N hammer head is stopped from an initial downward velocity of 3.2 m/s in a distance of 0.45 cm by a nail in a pine board. In addition to its weight, there is a 15-N downward force on the hammer head applied by the person using the hammer. Assume that the acceleration of the hammer head is constant while it is in contact with the nail and moving downward.

2. Relevant equations
F=ma
kinematics

3. The attempt at a solution
I know a few things....that the nail can be acting on the hammer with a normal force of 4.9N, the hammer is acting on the nail with a force of 15N, and the nail weighs 4.9N. There is most likely a constant acceleration of "a" acting on the system since we have a velocity and such....but when I solved for "a" with the given information with the equation of v^2 =(v_0)^2 +2a(x-x0) I got 1138m/s^2. I'm about 99% sure that is wrong. lol Please, help me wrap my head around some of the conceptual things. I get basic forces, but when it comes to difficult problems like this I'm lost as heck. Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 5, 2012

### tiny-tim

Welcome to PF!

Hi LoveHOPEpeace! Welcome to PF!
erm

what is the actual question?
seems ok

remember, this acceleration is up, but both the weight and the applied force are down … the main force is coming from the nail

3. Feb 5, 2012

### LoveHOPEpeace

Q- Calculate the downward force exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward.
Q2 - Suppose the nail is in hardwood and the distance the hammer head travels in coming to rest is only 0.12 cm. The downward forces on the hammer head are the same as on part (A). What then is the force exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward?

I fail! I never posted the questions.... lmao
Just a setup and an understanding of why you approached the problem that way would be very helpful. Thanks!

4. Feb 5, 2012

### tiny-tim

Your a (acceleration) seems to be correct.

Now use good ol' Newton's second law (Ftotal = ma) to find the total force, then subtract the known force to find the unkown one.