Final answer:
The normal force on each of the bodybuilder's feet while using a leg apparatus with a resistance of 1020N would be 510N if the force is equally distributed. This calculation assumes the system is in static equilibrium.
Step-by-step explanation:
To determine the normal force on each of the bodybuilder's feet when she pushes a leg apparatus with a resistance of 1020N, we can assume that the system is in static equilibrium because she comes to rest at full extension. In static equilibrium, the sum of the forces is zero. The apparatus exerts a force downward, and the bodybuilder's legs exert an equal force upward. This upward force is transmitted through her feet to the ground, which provides the normal force.
If the bodybuilder uses both feet equally to exert force on the leg apparatus, then the normal force on each foot would be half of the total resistance force. Therefore, the normal force on each foot would be 1020N / 2 = 510N.
It is important to note that this is an idealization and the actual distribution of force can vary based on the bodybuilder's stance and the mechanics of the leg press machine. Additionally, if the bodybuilder weighs more than the force exerted, her weight would also contribute to the normal force. If evenly distributed, the combined normal force from her weight and the resistance would be shared equally between her two feet.