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As a 5.0 × 10^2-newton basketball player jumps

from the floor up toward the basket, the

magnitude of the force of her feet on the floor is

1.0 × 10^3 newtons. As she jumps, the magnitude

of the force of the floor on her feet is

(1) 5.0 × 10^2 N (3) 1.5 × 10^3 N

(2) 1.0 × 10^3 N (4) 5.0 × 10^5 N

Verified Answer :

This player is initially at rest, then the Force Weight (W) and the Normal Force (N) have a same module. When applies a Extra Force (E) on the floor, this reacts with this increase, causing the player to be released upwards.
     Using the Newton's Secound Law, we have:

\(F_{R}=ma \\ W+E-N=ma \\ N =W+E \\ N=5*10^2+10*10^2 \\ \boxed {N=1.5*10^3N}\)

Number 3

If you notice any mistake in my english, please let me know, because i am not native.

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