Answer:
[tex]14.68\ \text{m}[/tex]
Masses of the truck and load
Explanation:
v = Final velocity = 0
u = Initial velocity = 12 m/s
[tex]\mu[/tex] = Coefficient of friction = 0.5
a = Acceleration
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
s = Displacement
f = Friction
F = Force applied
The force balance of the system is given by
[tex]F=-f\\\Rightarrow ma=-\mu mg\\\Rightarrow a=-\mu g[/tex]
From the kinematic equations we have
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2(-\mu g)}\\\Rightarrow s=\dfrac{0-12^2}{2(-0.5\times 9.81)}\\\Rightarrow s=14.68\ \text{m}[/tex]
The minimum stopping distance for which the load will not slide forward relative to the truck is [tex]14.68\ \text{m}[/tex]
As it can be seen that the masses of the truck and load are not used in the above used formulae. So, they are unnecessary.
