A 1260 -kg car is pushing an out-of-gear 2180 -kgtruck that has a dead battery.

Question

A 1260 -kg car is pushing an out-of-gear 2180 -kgtruck 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 4520 N . The rolling friction of the car can be neglected, but the heavier truck has a rolling friction of 760 N , including the "friction" of turning the truck's drivetrain. What is the magnitude of the force the car applies to the truck?

What is the magnitude FConT of the force that the car exerts on the truck? (Answer in Newtons)

Explanation / Answer

F = ma

a = F/m

a = (4520 - 760) / ( 1260 + 2180)

a = 1.093 m/s^2

FConT - 760 = 2180*1.093

Fcont = 3142.79

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