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?

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

ap8997154 ap8997154 · Answer 2 · 2022-03-02T08:33:30+01:00

Friction force is the product of the coefficient of friction and the normal force. The magnitude of the force the car applies on the truck is 3650 N.

What is friction force?

The friction force is the force exerted by the surface as the object moves across it or makes an effort to move across it. It is given by the formula,

[tex]F_r = \mu N[/tex]

where N is the normal force.

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

[tex]\rm F_t = F-f_r\\\\F_t = 4420-770\\\\F_t = 3650 N[/tex]

Hence, the magnitude of the force the car applies on the truck is 3650 N.

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