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 = 10000 kg
mass of flat bed = 20000 kg
initial speed of truck = 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 -------------let this be equation 1
where = normal force = mg
so
Fs,max = μs mg
ma = μs mg
divide through by mass
a = μs g ---------- let this be equation 2
in equation 2, we substitute in our values
a = 0.5 × 9.8 m/s²
a = 4.9 m/s²
now, from the third equation of motion
v² = u² + 2as
² =
² + 2aΔx
where 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
