Air Resistance of a Diver using Ek and Eg

[physicsmyfav]

Homework Statement

A 57.0 kg diver dives from a height of 15.0 m. She reaches a speed of 14.0 m/s just before entering the water. What was the average force of air resistance (e.g., friction) acting on the diver?
What is the force of friction underwater if she reaches a depth of 2.5 m before stopping? Do not neglect the buoyant force of 500 N acting on the diver once underwater

Homework Equations

Eg=mgh
Ek=(1/2)mv^2
Et=Ek+Eg
W=Ef-Ei
W=change in Ek
W=fd

The Attempt at a Solution

Ek=(1/2)(57kg)(14m/s)^2
=5586J

W=Ek
=5586J-0J

W=fd
5586J=F(15m)
5586/15=F
F=372.4N

Fnet=ma
=(57kg)(9.8N/kg)
=558.6N

Fnet=558.6N
Fapp-Ff=558.6N
372.4N-Ff=558.6N
Ff=186.2N

I'm not sure if I solved for friction properly. The way that I solved here doesn't work for the next step in the water, so I think initially started wrong.

 

[FermiAged]

How does the actual velocity of the diver about to enter the water compare to what would be expected in a vacuum? What would that imply about the actual acceleration?

 

[physicsmyfav]

The only acceleration would be that of gravity 9.8.

 

[FermiAged]

The only acceleration would be that of gravity 9.8.

What I am trying to say is that the velocity of the diver ( as given in the problem) is somewhat less than what would result from g when falling from the height (as given in the problem). Clearly the diver is accelerating less than g and that is due to the force of air resistance. You can use Newton's Second law to determine what this force is.

 

[FermiAged]

It just occurred to me that the problem or the instructor wants the problem solved by COE. In this case, unless I made a math mistake, I get an answer close to yours. I took the sum of forces approach because it was a more straightforward way of addressing the buoyant force once the diver enters the water.