How Does Wood Slow Down a Bullet?

[mandabox]

Homework Statement

A bullet of mass 0.0022 kg moving at 488 m/s embeds itself in a large fixed piece of wood and travels 0.56 m before coming to rest. Assume that the deceleration of the bullet is constant.

What force is exerted by the wood on the bullet?
F = N
1.01 NO

HELP: Find the deceleration of the bullet.
HELP: Use Newton's Second Law

Homework Equations

velocity & displacement:
v²=v₀²+2aΔx

Newton's 2^nd Law:
F⃗_net = ΣF⃗ = ma⃗

The Attempt at a Solution

Well clearly in the help options it told me to do what I had already done based on instinct.
I plugged v = 0 , v₀ = 488m/s , and Δx = 0.56 m. I figured that at the end of 0.56 m the final velocity would have to be 0 and obviously when the bullet begins its deceleration it had hit the wood. So I plugged in 0² = (488m/s)² + 2a(0.56m) and came out with the result a = -461.116657m/s².

I plugged this into the formula for force and got an answer of -1.01. I tried both -1.01 and +1.01 and neither attempt worked and I got the same answer no matter the number of sigfigs I did my calculations with. Am I getting it wrong because I've overlooked something? This problem seemed easy, what am I doing wrong? Doesn't the wood exert a force that's equal and opposite on the force of the bullet? I am assuming the bullet is traveling parallel to the surface of the Earth along the positive x-axis.

 

[mandabox]

I just went even simpler and used v = Δd/Δt to find time and then a = Δv/Δt to find acceleration. But the time to the last decimal shown on my calculator is 0.001147541 seconds and the acceleration is an enormous number in the four hundred thousands! And depending on how many sigfigs I take from time dictates how large or small my acceleration will be and I've gotten a force ranging from 930 to 976 and none of them seem right. I've dealt with this TYCHO system before and I KNOW that it is picky about your numbers. But whether I keep all the numbers to the end or round it off to two sigfigs at EVERY step, it's wrong! I'm getting really frustrated with this.

 

[mandabox]

After much deliberation and calculations I've discovered that the formula to use is in Chapter 8 of my textbook. 1/2mv²=Fd.

A bullet of mass 0.0022 kg moving at 488 m/s embeds itself in a large fixed piece of wood and travels 0.56 m before coming to rest. Assume that the deceleration of the bullet is constant.
What force is exerted by the wood on the bullet?
F = N *
467.78 OK

HELP: Find the deceleration of the bullet.
HELP: Use Newton's Second Law

I must have been plugging it into the original equations wrong because they also worked after I did a reverse calculation. I'm not sure how someone can make the same error for an hour but I guess that's what happened.

 

[Shootertrex]

You should be able to easily use the Work-Energy Theorem for this problem:

$\Sigma W=\Delta KE$

You know the change in kinetic energy, and you know the distance that the force is applied over.

 

[asteriatic]

Your calculations for the deceleration and the resulting force are correct. However, the negative sign indicates that the force is acting in the opposite direction of the initial velocity of the bullet, which is towards the wood. This makes sense because the wood is exerting a force to stop the bullet's motion.

In order to get the correct answer, you need to pay attention to the directions of the forces. In this case, the force exerted by the wood on the bullet is in the opposite direction of the bullet's initial velocity. So the correct answer would be F = -1.01 N. The negative sign indicates that the force is acting in the opposite direction of the initial velocity.

In general, when dealing with forces, it is important to pay attention to the directions of the forces and to assign negative signs to forces that act in the opposite direction of the chosen positive direction. This will help you get the correct answer.