Asked by Ntshade68
A 1230 -kg car is pushing an out-of-gear 2180 -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 4420N . The rolling friction of the car can be neglected, but the heavier truck has a rolling friction of 750N , including the "friction" of turning the truck's drivetrain. What is the magnitude of the force the car applies to the truck?
Answers
Answered by
Damon
The force from the car on the truck = 4420 - 1230 a
So our task is to find a
total m = 1230 + 2180 = 3410 kg
total F = 4420 - 750 = 3670 N
so
a = 3670/3410 = 1.08 m/s^2
so finally
F on truck = 4420 - 1230 (1.08)
F on truck = 3096 Newtons
So our task is to find a
total m = 1230 + 2180 = 3410 kg
total F = 4420 - 750 = 3670 N
so
a = 3670/3410 = 1.08 m/s^2
so finally
F on truck = 4420 - 1230 (1.08)
F on truck = 3096 Newtons
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