Answer:
Force, [tex]F=2.27\times 10^4\ N[/tex]
Explanation:
Given that,
Mass of the bullet, m = 4.79 g = 0.00479 kg
Initial speed of the bullet, u = 642.3 m/s
Distance, d = 4.35 cm = 0.0435 m
To find,
The magnitude of force required to stop the bullet.
Solution,
The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :
[tex]F.d=\dfrac{1}{2}m(v^2-u^2)[/tex]
Finally, it stops, v = 0
[tex]F.d=-\dfrac{1}{2}m(u^2)[/tex]
[tex]F=\dfrac{-mu^2}{2d}[/tex]
[tex]F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}[/tex]
F = -22713.92 N
[tex]F=2.27\times 10^4\ N[/tex]
So, the magnitude of the force that stops the bullet is [tex]2.27\times 10^4\ N[/tex]
