Question

A 54.0 kg woman steps off of a diving platform with a height of 12.0 m and drops straight down into the water. If she reaches a depth of 4.9 m, what is the average magnitude of the resistance force F average exerted on her by the water

Answer 1

Answer:

1296 N

Explanation:

First of all, we calculate what is the initial mechanical energy of the woman: when she stands on the diving platform, all her energy is gravitational potential energy, so

$E=GPE=mgh$

where

m = 54.0 kg is the mass of the woman

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 12.0 m is the initial height

Therefore,

$E=(54.0)(9.8)(12.0)=6350 J$

Assming there is no air resistance, this mechanical energy remains conserved until the woman reaches the water. Then, after that, all her energy is dissipated through a distance of

d = 4.9 m

Due to the presence of the resistive force F. Since energy is conserved, this means that the work done by this resistive force is equal (in magnitude) to the initial energy of the woman, so:

$W=E$

The work done by the resistive force can be written as

$W=Fd$

Therefore combining the two equations, we find the average magnitude of the force:

$F=\frac{E}{d}=\frac{6350}{4.9}=1296 N$