# Energy lost due to dissipation by air drag

1. May 5, 2008

### rosstheboss23

1. The problem statement, all variables and given/known data
An outfielder throws a baseball with an initial speed of 71.0 mi/h. Just before an infielder catches the ball at the same level, its speed is 85 ft/s. How much of the ball's mechanical energy is dissipated by air drag? (The weight of a baseball is 9 oz.) Answer is asked for in ft lb

2. Relevant equations
KE=KE+Energy lost to air drag 1/2mv^2= KE 5280ft=1mi 1h=3600s 1J=.738ft lb

3. The attempt at a solution
I converted 71mi/hr to m/s and got 31.7479m/s then I converted 85ft/s to m/s and got 25.91m/s. I then used KE= KE + energy lost to air drag and canceled out masses and got 168.18J= energy lost to drag. I then converted using the books conversion and got 124.12 ft lb. I have no idea where I when wrong. Thanks whoever helps.

2. May 5, 2008

### alphysicist

Hi rosstheboss23,

I don't think you can cancel out the masses; you don't have an explicit mass term in the energy loss term.

3. May 5, 2008

### rosstheboss23

Ok. If I can't cancel the masses then what should I do to make progress on this problem. The ball is at the same height so I assumed that I could use conservation of energy and just add loss of energy due to drag to the final KE. (Based on the same height I eliminated change in PE because it is the same.

4. May 5, 2008

### alphysicist

I'm not sure what your question is. You already have written the expression you need:

$$KE_i = KE_f + (\mbox{energy dissipated due to air drag})$$

You also have the formula for kinetic energy (1/2 mv^2); you have the initial and final speeds and the mass of the ball (which you can find from the weight), so you can go ahead and calculate the energy dissipated. There seems to be a lot of unit conversions (if you want to do the calculations in SI units), but I think you already have everything you need.

5. May 5, 2008

### rosstheboss23

Ok. Thanks. I see what I did wrong...I thought the masses canceled. Thanks again. I appreciate it.

6. Sep 22, 2008

### jakew1987

I have a similar problem that I am working on. In this case, however, I am being asked to find the energy dissipated by performing a line integral on the force of air resistance. The object is spherical with a diameter of .0732m, which best suits a quadratic approximation (F=C2*V^2). The problem is V is a function of time (or position, depending on how you look at it). So right now I have the work done on the ball by the air, W = INT[f(v)] dv = C2*INT[(VT^2)*(1-e^(-2gy/(VT^2)))] dv. Note that the integrand contains the variable y. Using change of base therom, I could come up with a pretty ugly looking integrand that I really don't want to compute, but I was wondering if anyone knew of a better way to go about this.