Answer:
a) F = 2.3797 10⁴ N, b) t = 1.97 10⁻⁴ s
Explanation:
a) For this exercise let's use the relationship between work and scientific energy
W = ΔK
W = F .d
ΔK = ½ m v² - ½ m v₀²
the bold are vectors. in this case the force of the bullet on the hand and the displacement of the bullet has the opposite direction, therefore the angle between the two is 180. The velocity when the bullet stops is zero.
We substitute
-F x = - ½ m vo2
F = [tex]\frac{m v_o^2}{2 x}[/tex]
let's calculate
F = [tex]\frac{7.80 \ 10^{-3} 600^{2} }{2 \ 5.90 \ 10^{-2} }[/tex]
F = 2.3797 10⁴ N
b) let's find the acceleration
F = ma
a = F / m
a = 2.3797 10⁴ / 7.80 10⁻³
a = 3.05 10⁶ m / s²
now we can use the kinematics relations
v = v₀ - a t
the final velocity is zero v = 0
t = v₀ / a
t = 600 / 3.05 10⁶
t = 1.97 10⁻⁴ s
