1. The problem statement, all variables and given/known data 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 2. Relevant equations velocity & displacement: v2=v02+2aΔx Newton's 2nd Law: F⃗net = ΣF⃗ = ma⃗ 3. 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.