Question

A 2300 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full power and good tires on good pavement, the maximum forward force on the truck is 18,000 N.

Part A

What is the maximum possible acceleration the truck can give the SUV?

Express your answer to two significant figures and include the appropriate units.

a =

Part B

At this acceleration, what is the force of the SUV's bumper on the truck's bumper?

Express your answer to two significant figures and include the appropriate units.

Fsuv on truck =

Answer 1

A) the maximum possible acceleration = 3.83 m/s²

B) F(SUV bumper) = 2400 kg *3.83 m/s² = 9192 N = 92 * 10² N

Step-by-step 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

Answer 2

a)
`a_(max) = 3.8\,(m)/(s^(2))`, b)
`F = 9260\,N`

Step-by-step explanation:

a) The maximum possible acceleration that the truck can give to the SUV is:

`a_(max) = (18,000\,N)/(2,400\,kg+2300\,kg)`

`a_(max) = 3.8\,(m)/(s^(2))`

b) The equation of equilibrium for the truck is:

`18,000\,N - F = (2,300\,kg)\cdot (3.8\,(m)/(s^(2)))`

The force of the SUV's bumper on the truck's bumper is:

`F = 9260\,N`