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

"KE 1= 1/2*0.002kg*300^2m/s=30 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...)

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 ?

To find the average force during the impact, we can use the conservation of energy principle. Initially, the bullet has kinetic energy (KE) given by:

KE1 = 1/2 * m * v1^2

where m is the mass of the bullet (0.002 kg) and v1 is its initial velocity (3.00 x 10^4 cm/s converted to meters per second). Therefore, substituting the values:

KE1 = 1/2 * 0.002 kg * (30000 m/s)^2 = 30 J

As the bullet comes to rest, its final kinetic energy (KE2) becomes zero. This means all its initial kinetic energy is converted into work done against the tree stump. So, the work done by the bullet on the tree is also 30 J.

Now, using the work-energy theorem, we can relate the work done to the force applied and the distance penetrated:

Work = force * distance

Rearranging the equation, we have:

force = work / distance

Plugging in the values, we get:

force = 30 J / 0.05 m = 600 N

Therefore, the average force during the impact is 600 Newtons.