Answer:
[tex]23796.19\ \text{N}[/tex]
[tex]0.0002032\ \text{s}[/tex]
Explanation:
F = Force
s = Displacement = 6.3 cm
m = Mass of bullet = 7.8 g
v = Velocity of bullet = 620 m/s
t = Time taken
Work done is given by
[tex]W=Fs[/tex]
Kinetic energy is given by
[tex]K=\dfrac{1}{2}mv^2[/tex]
Using work energy considerations we get
[tex]Fs=\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{1}{2s}mv^2\\\Rightarrow F=\dfrac{1}{2\times 0.063}\times 7.8\times 10^{-3}\times 620^2\\\Rightarrow F=23796.19\ \text{N}[/tex]
The average force that stops the bullet is [tex]23796.19\ \text{N}[/tex].
Force is given by
[tex]F=m\dfrac{v-u}{t}\\\Rightarrow t=m\dfrac{v-u}{F}\\\Rightarrow t=7.8\times 10^{-3}\times \dfrac{620}{23796.19}\\\Rightarrow t=0.0002032\ \text{s}[/tex]
The time taken to stop the bullet is [tex]0.0002032\ \text{s}[/tex].
