Question

a 2.1 gram bullet is shot into a tree stump. Just before it hits the stump the bullet is moving at 280 m/s. The bullet penetrates into the tree and comes to a rest after having gone 5.1 cm into the tree

a.What was the initial kinetic energy of the bullet

b.What is the final kinetic energy of the bullet

c.How much work was done on the tree

d.what was the average force during the impact

Answer 1

a, b.
KE = (1/2)m*V^2

c.
The work done is the change in kinetic energy.

d.
W = F*d