Answer:
a) [tex]a_{max} = 3.8\,\frac{m}{s^{2}}[/tex], b) [tex]F = 9260\,N[/tex]
Explanation:
a) The maximum possible acceleration that the truck can give to the SUV is:
[tex]a_{max} = \frac{18,000\,N}{2,400\,kg+2300\,kg}[/tex]
[tex]a_{max} = 3.8\,\frac{m}{s^{2}}[/tex]
b) The equation of equilibrium for the truck is:
[tex]18,000\,N - F = (2,300\,kg)\cdot (3.8\,\frac{m}{s^{2}})[/tex]
The force of the SUV's bumper on the truck's bumper is:
[tex]F = 9260\,N[/tex]
Answer:
A) the maximum possible acceleration = 3.83 m/s²
B) F(SUV bumper) = 2400 kg *3.83 m/s² = 9192 N = 92 * 10² N
Explanation:
Step 1: Data given
Mass of the truck = 2300 kg
Mass of the SUV = 2400 kg
The maximum forward force on the truck is 18000 N
Step 2: What is the maximum possible acceleration the truck can give the SUV?
Total mass = MAss truck + mass SUV
Total mass = 2300 kg + 2400 kg
Total mass = 4700 kg
Acceleration = Force / mass
The maximum possible acceleration = 18000 N / 4700 kg
the maximum possible acceleration = 3.83 m/s²
Step 3: At this acceleration, what is the force of the SUV's bumper on the truck's bumper?
F = m * a
F(trucks bumper) = 2300 kg *3.83 m/s² = 8809 N
F(SUV bumper) = 2400 kg *3.83 m/s² = 9192 N = 92 * 10² N
