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

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

Answer 2

Answer:

a) $a_{max} = 3.8\,\frac{m}{s^{2}}$, b) $F = 9260\,N$

Explanation:

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

$a_{max} = \frac{18,000\,N}{2,400\,kg+2300\,kg}$

$a_{max} = 3.8\,\frac{m}{s^{2}}$

b) The equation of equilibrium for the truck is:

$18,000\,N - F = (2,300\,kg)\cdot (3.8\,\frac{m}{s^{2}})$

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

$F = 9260\,N$