Work done by air on a baseball

[Chandasouk]

Homework Statement

The following is data from a computer simulation for a batted baseball with mass 0.145 kg, including air resistance:
t x y v_x v_y
0 0 0 30.0 m/s 40.0 m/s
3.05 s 70.2 m 53.6 m 18.6 m/s 0
6.59 s 124.4 m 0 11.9 m/s -28.7 m/s

How much work was done by the air on the baseball as it moved from its initial position to its maximum height?

How much work was done by the air on the baseball as it moved from its maximum height back to the starting elevation?


Man, I am clueless about this problem. The baseball has to move in a parabolic motion though, right? It's a batted ball.

I made a bar graph first of all


(Initial) KE = PE+KE(Final)

Initially, the Vx and Vy are 30.0 m/s and 40.0 m/s respectively. I did the Pythagorean theorem and obtained V = 8.37m/s

Now

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

.5(.145kg)(8.37m/s)^2 = (.145kg)(9.80m/s^2)(53.6m) +1/2(.145kg)(18.6m/s)2 + $$Delta$$U_internal

-96.17J = U_internal

?

 

[PhanthomJay]

Homework Statement

The following is data from a computer simulation for a batted baseball with mass 0.145 kg, including air resistance:
t x y v_x v_y
0 0 0 30.0 m/s 40.0 m/s
3.05 s 70.2 m 53.6 m 18.6 m/s 0
6.59 s 124.4 m 0 11.9 m/s -28.7 m/s

How much work was done by the air on the baseball as it moved from its initial position to its maximum height?

How much work was done by the air on the baseball as it moved from its maximum height back to the starting elevation?


Man, I am clueless about this problem. The baseball has to move in a parabolic motion though, right? It's a batted ball.

right.

I made a bar graph first of all


(Initial) KE = PE+KE(Final)

you forgot to include the work done by the air in this equation.

Initially, the Vx and Vy are 30.0 m/s and 40.0 m/s respectively. I did the Pythagorean theorem and obtained V = 8.37m/s

try it again, you have an error here, it's v^2 = vx^2 + vy^2

Now

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

.5(.145kg)(8.37m/s)^2 = (.145kg)(9.80m/s^2)(53.6m) +1/2(.145kg)(18.6m/s)2 + $$Delta$$U_internal

-96.17J = U_internal

?

correct your math error for v, and you should be OK. How about the downward journey? Watch your plus and minus signs.

 

[Chandasouk]

Where does the work done by air go to? Initial? How would it be expressed in an equation?

 

[PhanthomJay]

Where does the work done by air go to? Initial? How would it be expressed in an equation?

you will end up with the correct answer in the end when you correct the math, but your equations left off the work term or U_therm term initially, then you added it in later.
There are several ways to write the general Conservation of Energy equation. One of them is

$$KE_i + PE_i + W_{nc} = KE_f + PE_f$$ where $$W_{nc}$$ is the work done by all nonconservative forces like air resistance, friction, applied forces, etc. Another way is to write it

$$KE_i + PE_i + W_{applied forces} = KE_f + PE_f + U_{therm}$$ where $$U_{therm}$$ is the internal heat energy caused by the the work done by friction, air resistance, or other heat producing non-conservative forces. I like the first one better. There are other ways, like

$$W_{nc} = \Delta KE + \Delta PE$$, which is the same as the first equation in another form. As always, be sure to mind your plus and minus signs.

 

[Chandasouk]

Doing the math correctly now


.5(.145kg)(50m/s/s)^2 = (.145kg)(9.80m/s^2)(53.6m) +1/2(.145kg)(18.6m/s)2 + LaTeX Code: Delta Uinternal

181.25J=101.2477J + $$\Delta$$U_internal

$$\Delta$$U_internal = 80.0023J

But my homework accepts it at -80.0023J. I'm guessing it is because friction always opposes so it would be negative work?


For the second part of the question: How much work was done by the air on the baseball as it moved from its maximum height back to the starting elevation?

Initial (Top) KE + PE = Final (Bottom) KE+ U_internal

101.2477J = .5(.145kg)(31.069m/s)^2 + U_internal

I got the 31.069m/s by

(11.96)^2+(28.7)^2=V^2

101.2477J = 69.98300017J + U_internal

31.26J = U_internal ?

 

[PhanthomJay]

Doing the math correctly now


.5(.145kg)(50m/s/s)^2 = (.145kg)(9.80m/s^2)(53.6m) +1/2(.145kg)(18.6m/s)2 + LaTeX Code: Delta Uinternal

181.25J=101.2477J + $$\Delta$$U_internal

$$\Delta$$U_internal = 80.0023J

But my homework accepts it at -80.0023J. I'm guessing it is because friction always opposes so it would be negative work?

yes, that's one reason why i like the other form of the equation that uses work instead of thermal or internal energy; the minus sign takes care of itself that way.

For the second part of the question: How much work was done by the air on the baseball as it moved from its maximum height back to the starting elevation?

Initial (Top) KE + PE = Final (Bottom) KE+ U_internal

101.2477J = .5(.145kg)(31.069m/s)^2 + U_internal

I got the 31.069m/s by

(11.96)^2+(28.7)^2=V^2

101.2477J = 69.98300017J + U_internal

31.26J = U_internal ?

Oh there you go again, right number, wrong sign. What happened to the minus sign? The work done is negative, but i guess you consider the internal energy as positive, but the problem asks for the work, not energy.