Answer:
Force(F) = -80,955.01 N
Explanation:
We need to first determine the impulse that the truck driver received from the car during the collision
So; m₁v₁ - m₂v₂ = (m₁m₂)v₀
where;
m₁ = mass of the truck = 4280 kg
v₁ = v₂ = speed of the each vehicle = 7.69 m/s
m₂ = mass of the car = 810 kg
Substituting our data; we have:
(4280×7.69) - (810×7.69) = (4280+810)v₀
32913.2 - 6228.9 = (5090)v₀
26684.1 = (5090)v₀
v₀ = 
v₀ = 5.25 m/s
NOW, Impulse on the truck = m (v₀ - v)
= 4280 × (5.25 - 7.69)
= 4280 × (-2.44)
= -10,443.2 kg. m/s
Force that the seat belt exert on the truck driver can be calculated as:
Impulse = Force × Time
-10,443.2 kg. m/s = F (0.129)
F = 
Force(F) = -80,955.01 N
Thus, the Force that the seat belt exert on the truck driver = -80,955.01 N
The answer is going be desert.
Answer:
v=0.94 m/s
Explanation:
Given that
M= 5.67 kg
k= 150 N/m
m=1 kg
μ = 0.45
The maximum acceleration of upper block can be μ g.
a= μ g ( g = 10 m/s²)
The maximum acceleration of system will ω²X.
ω = natural frequency
X=maximum displacement
For top stop slipping
μ g =ω²X
We know for spring mass system natural frequency given as

By putting the values

ω = 4.47 rad/s
μ g =ω²X
By putting the values
0.45 x 10 = 4.47² X
X = 0.2 m
From energy conservation


150 x 0.2²=6.67 v²
v=0.94 m/s
This is the maximum speed of system.