Having some trouble ie. help me (Newton's Laws of Motion - Basic stuff)

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SUMMARY

The discussion focuses on calculating the downward force exerted by a hammer head on a nail while in contact, given specific parameters. A 4.9N hammer head, with an additional 15N force applied by the user, is stopped from an initial velocity of 3.2 m/s over a distance of 0.45 cm. The correct calculation involves determining the acceleration and mass of the hammer head, leading to a downward force of 569N, which includes the applied force. The confusion arises from differing interpretations of the calculations, with some participants referencing a textbook answer of 590N.

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Supaiku
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A 4.9N hammer head is stopped from an inital downward velocity of 3.2m/s in a distance of .45cm by a nail in a pine board. In addition to its weight there is a 15N downward force on the hammer by the peson using the hammer. Assume the acceleration of hte hammer head is constance while its in contact with the nail and moving downward.
a) Free body Diagram for the hammer head and identify the reaction force to each force in the diagram.
b) Calculate th downward force of F exerted by the hammer head on the nail while the hammer head is in contact with the nal and mocing downward.


I can't seem to get the right answer (b - 5.9X10^2 N). If someone could help me out with the steps (like equations to use and any explinations you think might necissary) that would be cool.

I think I have the freebody diagram but if there is some easy way to explain it that would be nice too... the second part is more important though.
 
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is it stopped at the distance of 45 cm?
 
Simonnava: No, 0.45 cm!

Supaiku: Your "free body" diagram should show the two downward forces on the hammer (its weight, 4.9 N, and the 15N force given by the person) and the (unknown) upward force,from the nail, on the hammer head.

You can find that upward force in two ways:

1 (the hard way) Calling the acceleration a, the speed at any moment, t, is v= v0- at (I am taking a to be positive and putting the negative sign for the "slowing" acceleration in explicitely) and the distance moved is x= v0t- (a/2)t2. You know that v0= 3.2 m/s and that the total distance moved is 0.45 cm= 0.0045 m. You can solve 3.2t- (a/2)t2= 0.0045 to determine the time required to stop (it depends on a), then put that back into 3.2- (a/2)t2= 0 (since the speed is 0 at that time and solve for a.

2 (much simpler): As long as the acceleration is constant, the "average speed" is just the average of the initial and final speeds: here 3.2 m/s and 0 m/s so the average speed is 3.2/2= 1.6 m/s. At that speed, it takes t= 0.0045/1.6= 0.0028125 seconds to stop. The (constant) acceleration necessary to reduce the speed from 3.2 m/s to 0 in 0.0028125 seconds is 3.2/0.0028125= 1137 m/s2.

F= ma. Now that you know a, to finish you will need to find m, the mass of the hammerhead (4.9 N was its weight, not its mass).
 
ya, .45cm, like I said. If it was 45cm that would be a HUGE nail... more like a steak... for giant vampires! :p

Thanks, Halls Of Ivy:D

But weird, the book gave 5.9X10^2 N and I got 571 N (I used .0028 as the time when I solved it after glancing at your explanation). You think they'r just wrong? I can't seem to do anything with the numbers to get 590 N (like t=.002 (equals 800 N) or .003 (equals 533 N)).
 
I know this is a really old post, but I thought I would add my 2 cents just in case there are any students like me who come across this for THEIR homework and need clarification.

569 is right, but you can't forget to add the additional 15N that the person holding the hammer is applying.
 

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