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:
[tex]\frac{-b+-\sqrt{b^{2}-4ac } }{2a}[/tex]
with a=1.3*10^6, b=-530, c=5.4*10^-2
t=2*10^-4s
