Answer:
25 m
Explanation:
The maximum acceleration that the load can withstand without slipping out of the truck occurs when the net force on the load is equal to the maximum frictional force, therefore:
[tex]ma_{max} =-\mu N[/tex]
where
m = 10,000 kg is the mass of the load
[tex]a_{max}[/tex] is the maximum acceleration
[tex]\mu = 0.400[/tex] is the coefficient of friction
N is the reaction force of the truck on the load
Since the surface is horizontal, the reaction force is equal to the weight of the load, [tex]N=mg[/tex], so from the previous equation we get
[tex]a_{max} = -\mu g = -(0.400)(9.8)=-3.92 m/s^2[/tex]
where the sign is negative since the direction of the acceleration is opposite to the motion of the truck.
At the maximum acceleration, the stopping distance of the truck is the minimum. Therefore, we can use the SUVAT equation:
[tex]v^2-u^2 = 2a_{max} d[/tex]
where
v = 0 is the final velocity of the truck
u = 14.0 m/s is the initial velocity
d is the minimum stopping distance
Solving for d, we find
[tex]d=\frac{-u^2}{2a_{max}}=\frac{-(14)^2}{2(-3.92)}=25 m[/tex]
