A 900 kg car pushes a 1900 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4800 N.
(a) What is the magnitude of the force of the car on the truck?
(b) What is the magnitude of the force of the truck on the car?
The first answer posted by Damon is correct, but part B is not, since the problem is an action-reaction pair, the forces will be the same, thus part B is the same answer as part A
(a) Well, if the car is pushing the truck, it must mean that the two forces are equal in magnitude but opposite in direction. So, the magnitude of the force of the car on the truck is 4800 N.
(b) Now, let's not forget Newton's third law! If the car is pushing the truck with a force of 4800 N, then the truck must be pushing the car with the same magnitude but opposite in direction. So, the magnitude of the force of the truck on the car is also 4800 N.
You could say they're in a "forceful" relationship, pushing each other with the same intensity!
To answer these questions, we need to consider Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In other words, the force exerted by one object on another is always equal in magnitude and opposite in direction to the force exerted by the second object on the first.
(a) What is the magnitude of the force of the car on the truck?
Since the car is pushing the truck, the force exerted by the car on the truck is equal to the force exerted by the drive wheels of the car on the ground. This force is given as 4800 N.
(b) What is the magnitude of the force of the truck on the car?
By Newton's third law, the force exerted by the truck on the car is equal in magnitude but opposite in direction to the force exerted by the car on the truck. Therefore, the magnitude of the force exerted by the truck on the car is also 4800 N. This force acts in the opposite direction to the force exerted by the car.
So, the magnitude of the force of the car on the truck is 4800 N, and the magnitude of the force of the truck on the car is also 4800 N.
The sled dog in figure drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10.
F = m a
4800 = (900+1900) a
a = 4800/2800 = 1.714 m/s^2
a)
F = m a
F = (1900)(1.714)
= 3257 Newtons
b)
negative 3257 Newtons (action and reaction)