Forces of a Hammer Acting On a Nail With constant acceleration

[LoveHOPEpeace]

Homework Statement

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.

Homework Equations

F=ma
kinematics

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!

 

[tiny-tim]

Welcome to PF!

Hi LoveHOPEpeace! Welcome to PF!

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.

erm …

what is the actual question? ​

… v^2 =(v_0)^2 +2a(x-x0) I got 1138m/s^2.

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

 

[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!

 

[tiny-tim]

Your a (acceleration) seems to be correct.

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

 

[asteriatic]

Firstly, it is important to note that the acceleration of the hammer head will not be constant throughout its entire motion. The acceleration will only be constant while the hammer is in contact with the nail and moving downward. Once the hammer comes to a stop, the acceleration will no longer be constant.

To solve this problem, we can use the equation F=ma, where F is the net force acting on the hammer, m is the mass of the hammer, and a is the acceleration. In this case, the net force acting on the hammer is the sum of the weight of the hammer (4.9N) and the downward force applied by the person (15N). Therefore, the net force is 19.9N.

Next, we can use the equation v^2=(v_0)^2+2a(x-x_0), where v is the final velocity, v_0 is the initial velocity (3.2 m/s), a is the acceleration, x is the final position (0.45 cm = 0.0045 m), and x_0 is the initial position (0 m). Solving for a, we get a = -1133.33 m/s^2.

This negative value for acceleration indicates that the hammer is actually decelerating, or slowing down, as it comes into contact with the nail. This makes sense since the nail is acting as a resistance force to the motion of the hammer.

In conclusion, the forces acting on the hammer are the weight of the hammer, the downward force applied by the person, and the normal force from the nail. The acceleration of the hammer will be constant while it is in contact with the nail and moving downward, but will change once it comes to a stop.