Two dogs are playing tug-o-war. The first dog pulls with the force of 400N and the second dog is hanging on but is being pulled towards the first dog at a constant velocity. What is the force of friction between the second dog's feet and the ground?
400-Fk = M*a.
400-Fk = M*0 = 0
Fk = 400 N.
To determine the force of friction between the second dog's feet and the ground, we need to understand the forces acting on the second dog. In this scenario, we have two opposing forces: the force exerted by the first dog and the force of friction.
1. The force exerted by the first dog: The first dog pulls with a force of 400N. This force is acting in one direction.
2. The force of friction: The force of friction acts in the opposite direction to the motion and prevents the second dog from being dragged by the first dog. Since the second dog is moving at a constant velocity, we know that the force of friction is equal to the force exerted by the first dog.
Therefore, the force of friction between the second dog's feet and the ground is also 400N.