Air Resistance of a Diver using Ek and Eg

1. Apr 23, 2014

physicsmyfav

1. The problem statement, all variables and given/known data
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

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

3. 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.

2. Apr 24, 2014

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?

3. Apr 24, 2014

physicsmyfav

The only acceleration would be that of gravity 9.8.

4. Apr 24, 2014

FermiAged

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.

5. Apr 24, 2014

FermiAged

It just occured 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.