A1300 -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?

Ft = 3650 N

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

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hope this you out ^w^