Question

A 2.0-gram bullet is shot into a tree stump. It enters at a speed of 3.00 x 104 cm/s and comes to rest after having penetrated 0.05 m in to the stump. What was the average force during the impact? Show all calculations leading to an answer.

KE 1= 1/2*0.002kg*300^2m/s=30 j KE2=0 work
Tree=30 j f=ma first find a V2^2-V1^2
= 2AX - 300^2M/s = 2*a*0.05 A=900,000m/sec^2 f =
0.002*9e5 = 1800 newtons

Answer 1

"KE 1= 1/2*0.002kg*300^2m/s=<i>30</i> j KE2=0 work "
should result in 90 J (and not 30 J), and assuming energy dissipation is uniform, then
KE=W=F*d, F=KE/d=90/.05=1800 N
which is exactly the answer you got for the second part.

Note: units which are proper names (pascals, newtons, joules...) are written in lower case when written in full, but in upper case when abbreviated (Pa, N, J...)

Answer 2

KE 1= 1/2*0.002kg*300^2m/s=30 j KE2=0 work
Tree=90 j f=ma first find a V2^2-V1^2
= 2AX - 300^2M/s = 2*a*0.05 A=900,000m/sec^2 f =
0.002*9e5 = 1800 newtons

so this would be correct
@mathmate ?