Question

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.

Answer 1

F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer

Answer 2

Acceleration = a = 3.75 m/s^2

Step-by-step explanation:

Mass of truck1 = m1 = 2300 kg

Mass of truck2 = m2 = 2500 kg

Total mass = m = m1 + m2 = 2300 + 2500 = 4800 kg

Force exerted by truck1 = F = 18000 N

As both trucks are joint together so, behaving as single object. The acceleration can be found by Newton’s second law of motion.

F = ma

a = F/m = 18000/4800

a = 3.75 m/s^2