Potential and Kinetic energy equations including drag coefficient

[ET333]

Homework Statement

I am currently writing a paper on conservation of energy in which I dropped a ball and calculated potential and kinetic energy. The issue I am having is that I have no idea how to incorporate drag force into my equations for P and K energy and also into my uncertainties. I have been looking every where for answers but cannot seem to find them. Please help!

Relevant Equations

Ke+Pe=Ke+Pe
Ke= 1/2 mv^2
Pe= mgh

1. Golf ball initial potential energy uncertainty (110-5kg 0.01 m 0.01m/s2)= 110-9J4.31210-5=4.31210-14j
2. Golf ball initial potential energy calculation (4.4 x10-4kg 9.8 m/s2 0.609 m)= 4.31210-54.31210-14j
3.Golf ball final potential energy uncertainty ( 110-5kg 0.0 m0.01 m/s2)= 0.0 J
4. Golf ball final potential energy calculation (4.4 x10-4kg 9.8 m/s2 0.0 m)= 0 0J
5. Initial velocity of golf ball Vi=dt → Vf=0.0m0.0 seconds =0m/s0J
6. Kinetic energy of golf ball initial calculation 1/2(4.4 x10-4kg0 m/s2)= 00 J
8. Final Kinetic energy of golf ball calculation 1/2(4.4 x10-4kg1.353 m/s2)= 4.0210-4 0.120 J
Final velocity of the golf ball uncertainty Vf uncertainty= 0.01m0.609m+0.01 seconds0.45 seconds=0.386 m/s
9. 1/2(4.4 x10-4kg1.353 m/s)= 4.0210-42.4710-5 J
10. Final kinetic energy uncertainty of the golf ball
((110-5kg4.410-4g)2+(20.0386m/s1.353m/s)2= 6.1410-2 x 4.0210-4 J=2.4710-5J
11. Initial Kinetic energy of golf ball calculation 1/2(4.4 x10-4kg0m/s2)=00 J
12. Initial kinetic energy uncertainty of the golf ball (110-5kg4.410-4g)2+(20.0m/s0.0m.s)2= 0.00 x 0.00 J=0.00J
13. Conservation of energy calculation
4.31210-54.31210-14J=4.0210-42.4710-5 J


14. Tennis ball initial potential energy uncertainty (110-5kg 0.01 m 0.01m/s2)= 110-9J3.2810-4J=3.2810-12j
15. Tennis ball initial potential energy calculation (5.5 x10-4kg 9.8 m/s2 0.609 m)=3.28210-33.2810-12J
16.Tennis ball final potential energy uncertainty ( 110-5kg 0.0 m0.01 m/s2)= 0.0 J
17. Tennis ball final potential energy calculation (5.5 x10-4kg 9.8 m/s2 0.0 m)= 0 0J
18. Initial velocity of golf ball Vi=dt → Vf=0.0m0.0 seconds =0m/s0J
Final velocity of golf ball Vi=dt → Vf=0.609m0.47 seconds =1.295 m/s
19. Kinetic energy of golf ball initial calculation 1/2(3.28 x10-4kg0 m/s2)= 00 J
20. Final Kinetic energy of golf ball calculation 1/2(5.5 x10-4kg(1.295 m/s)2)=4.6110-42.7610-5J
21. Final velocity of the tennis ball Vf = 0.609m0.47s=1.296m/s
22. Final velocity of the tennis ball uncertainty Vf uncertainty= 0.01m0.609m+0.01 seconds0.47 seconds=0.037 m/s

23. Final kinetic energy uncertainty of the tennis ball
((110-5kg5.510-4g)2+(20.037m/s1.29m/s)2= 5.9910-2 x 4.6110-4J=2.7610-5J
24. Initial Kinetic energy of tennis ball calculation 1/2(5.5 x10-4kg0m/s2)=00 J
25. Initial kinetic energy uncertainty of the tennis ball (110-5kg5.510-4g)2+(20.0m/s0.0m.s)2= 0.00 x 0.00 J=0.00J
26. Conservation of energy tennis ball drop
3.28210-33.2810-12J = 4.6110-42.7610-5J

 

[haruspex]

Drag on an accelerating object is a bit complicated. Starting from rest it is a linear function of speed, but at higher speeds transitions to quadratic. The transition point is determined by the Reynolds number. I am not sure where it would be in the context of your experiment.

If it is all in the linear phase then there is a closed form solution of the form $v(t)=v_{terminal}(1-e^{-\rho t})$.
There is no closed form solution to the quadratic form, but if the drop were far enough to achieve terminal velocity (it will not be for your data) then you could calculate the terminal velocity and deduce the work lost to drag.
Or you could model quadratic drag and get a numeric solution.