Answer:
A)
the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B)
data that were not necessary to the solution are;
a) mass of truck and b) mass of load
Explanation:
Given that;
mass of load [tex]m_{LS}[/tex] = 10000 kg
mass of flat bed [tex]m_{FB}[/tex] = 20000 kg
initial speed of truck [tex]v_{0}[/tex] = 12 m/s
coefficient of friction between the load sits and flat bed μs = 0.5
A) the minimum stopping distance for which the load will not slide forward relative to the truck.
Now, using the expression
Fs,max = μs [tex]F_{N}[/tex] -------------let this be equation 1
where [tex]F_{N}[/tex] = normal force = mg
so
Fs,max = μs mg
ma[tex]_{max}[/tex] = μs mg
divide through by mass
a[tex]_{max}[/tex] = μs g ---------- let this be equation 2
in equation 2, we substitute in our values
a[tex]_{max}[/tex] = 0.5 × 9.8 m/s²
a[tex]_{max}[/tex] = 4.9 m/s²
now, from the third equation of motion
v² = u² + 2as
[tex]v_{f}[/tex]² = [tex]v_{0}[/tex]² + 2aΔx
where [tex]v_{f}[/tex] is final velocity ( 0 m/s )
a is acceleration( - 4.9 m/s² )
so we substitute
(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx
0 = 144 m²/s² - 9.8 m/s²Δx
9.8 m/s²Δx = 144 m²/s²
Δx = 144 m²/s² / 9.8 m/s²
Δx = 14 m
Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B) data that were not necessary to the solution are;
a) mass of truck and b) mass of load
