Question

A 1300 -kg car is pushing an out-of-gear 2160 -kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push horizontally against the ground with a force of 4420 N . The rolling friction of the car can be neglected, but the heavier truck has a rolling friction of 770 N , including the "friction" of turning the truck's drivetrain. What is the magnitude of the force the car applies to the truck?

Answer 1

The magnitude of the force that the 1300-kg car applies to the 2160-kg truck, given a push force of 4420 N and a rolling friction of 770 N for the truck, is 3650 N.

Step-by-step explanation:

The scenario presented involves a 1300-kg car pushing a 2160-kg truck with a rolling friction of 770 N. The car exerts a horizontal force of 4420 N against the ground. To calculate the force the car applies to the truck, we can use Newton's Second Law of Motion by understanding that the net force applied on the truck by the car is the driving force minus the rolling friction force.

Net force on the truck = Force applied by the car - Rolling friction

Since we are given the total force exerted by the car's wheels and the rolling friction opposing the truck's motion, we can calculate the net force on the truck:

Net force on the truck = 4420 N - 770 N = 3650 N

Therefore, the magnitude of the force that the car applies to the truck is 3650 N.

Answer 2

Ft = 3650 N

Step-by-step explanation:

The car exerts a force on the truck and truck has a friction force, using Newton's second law we can find the net force is the difference between the two

Ft = F-fr

Ft = 4420-770

Ft = 3650 N

Directed from the car to a truck . The system acceleration is

Ft = (m₁ + m₂) a

a = Ft / (m₁ + m₂)

a = 3650 / (1300 +2160)

a = 1.05 m / s²