Question

A 7.80 g bullet moving at 530 m/s penetrates a tree trunk to a depth of 5.40 cm. (a) Use work and energy considerations to find the average frictional force that stops the bullet. (b) Assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving.

Answer 1

Answer:

a) F=20287.22N

b) t=2*10^-4s

Explanation:

E=1/2*m*v^2=0.5*7.8*10^-3kg*(530m/s)^2=1095.51J

The frictional force's work must be equal than the energy to stop the bullet.

So: W=F*d=F*0.054m=1095.51J, F=20287.22N

Considering the frictional force is constant, the bullet moves with constant aceleration.

a=F/m=20287.22N/7.8*10-2kg=2.6*10^6m/s^2

then d(t)=Vt-1/2*a*t^2,

5.4*10^-2m=530m/s*t-1.3*10^6m/s^2*t^2

I will calculate the time using the cuadratic formula:

$\frac{-b+-\sqrt{b^{2}-4ac } }{2a}$

with a=1.3*10^6, b=-530, c=5.4*10^-2

t=2*10^-4s