Final answer:
The net force required to stop a 1.2x10^4 kg truck traveling at 22 m/s in 330 m is approximately -876 N. This force is applied in the opposite direction to the truck's motion and is due to the frictional force produced by the truck's brakes.
Step-by-step explanation:
The question is asking for the net force required to bring a truck to a stop and the cause of this force. In physics, we use the principles of motion, specifically Newton's second law, to solve such problems.
a. We first need to find the truck's deceleration using the formula v^2 = u^2 + 2as, where 'v' is the final velocity (0 m/s in this case as the truck stops), 'u' is the initial velocity (22 m/s), 'a' is the acceleration (we need to find this), and 's' is the displacement (330 m). The deceleration (a negative form of acceleration) comes out to be -0.073 m/s^2. Then, we use Newton's second law F=ma where 'F' is the force, 'm' is the mass (1.2x10^4 kg), and 'a' is the acceleration to find the net force required. The force required to stop the truck in 330 m is approximately -876 N (negative indicating the force is applied opposite to the direction of motion).
b. This net force comes due to the frictional force applied by the brakes of the truck. The brakes create a force opposing the truck's motion, leading to its stop.
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