A 2300 kg truck has put its front bumper against the rear bumper of a 2500 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. What is the maximum possible acceleration the truck can give the SUV?

Acceleration a = 3.75 m/s^2

Explanation:

Given that

Mass M1 = mass of the truck = 2300kg

Mass M2 = mass of the SUV = 2500kg

Force applied F = 18000N

From Newton second law,

F = ma

Where a = acceleration

We must consider the entire mass involved in the system. Therefore,

F = (M1 + M2)a

Make a the subject of formula

a = F/(M1 + M2)

Acceleration a = 18000/( 2300 + 2500)

Acceleration a = 18000/ 4800

Acceleration a = 3.75 m/s^2

e) none of the above answers are correct

change tension