There will be two parts to this question.
1) the bullet strikes and stays inside the wood
2) both bullet and wood block travel as as a system for 9.9m, under the friction of 0.23.
Mass of bullet= Mb, Vb=speed of bullet
Mass of wood bloc=Mw, Vw= speed of wood
Mass of bullet and wood block=Ms, Vs= speed of system
1) MaVa+ MwVw= (Ma+Mw) Vs; Vw at rest initially.
1) Friction force= u.Ms.g= 0.23x1.376x9.8 =3.10N
Friction work= F.distance= 3.10N x 9.9m=30.7J
(we need to calculate the Kinetic energy before and after of the system, as it travels actoss 9.9m with friction.
1) KEinitial= KE final
Work friction+ KE initial = KE final
Wf + 0.5MsVsinital(squared) 0.5MsVsfinal
((Vs final is zero after 9.9m. So we're left))
Wf + O.5MsVsinitial(squared )=0
(plug the Vs initial answer from step 3 into the Vs in step 1)
=(1.376x 6.68/ 0.026kg
= 353.5m/s <- (ANSWER)
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