# AP Physics C Energy Problem! (Kinetmatics as well)

Discussion in 'Community Discussion' started by fivetoadsloth, Oct 13, 2009.

Joined:
Aug 15, 2006
#1
Here is the main question: the rest is just showing what I have tried. This is not actually a homework assignement, per se, and given the amount of energy I have put into it I'd really like to see it worked through.

An object of mass 1 Kg is launched at an angle of 42° from the horizontal at 180 m/s. A force of drag acts on it that is dependant on its velocity, with the force being equal to -.012V. What energy does the ball have when it reaches its maximum height?

Thanks a ton!

1. The problem statement, all variables and given/known data
First, this is not an assignment, per se, it is for my own help, and If someone could help me work it through that would be much appreciated.

An object of mass 1 Kg is launched at an angle of 42° from the horizontal at 180 m/s. A force of drag acts on it that is dependant on its velocity, with the force being equal to -.012V. What energy does the ball have when it reaches its maximum height?

2. Relevant equations
I solved the initial components in both the x and y directions and got 133.8 m/s and 120 m/s.
Other equations:
E=K+U
K=.5mv^2
U=mgh
x(t)=x+(integral)v(t)dt
I am sure I am missing some, which is part of my problem.

3. The attempt at a solution

I started with F=mdv/dt
Integrated to
Integral from Vi to V (dv')/fv'=t/m which didnt help much.
I then tried
mdv/dt=mg-kv (kv being my -.012v)
But, was having trouble with my signs, and it didnt look like it was coming out right.

THe retarding forces in the y direction are -mg-.012v
and the x direction is only -.012 v

Going to F=ma I tried various substitutions and again, had little luck.

I know I wiill need to solve the height at which the velociy is zero, using only the y component to get potential enrgy, and also find the x velcity at this time to get the kinetic energy and then add them up to get the total energy. The thing that is giving me difficult is the retarding force as a function of the velocity, so the force will be constantly changing.
I have been working on it, and got stuck, though it seems like this route would work.
Using F=ma and trying to solve for the downward force in the y direction I get
F=-mg-.012v
F=-mg-.012(dx/dt)
Integral of F dt= -mg-integral .12 dx
From the looks of this, I think that setting up the differential equation was not the right step, and am still unsure where to go from here.

2. ### Ttownbeast macrumors 65816

Joined:
May 10, 2009
#2
Based on the question you might be over thinking this--you are asking "what kind of energy or what amount" am I correct? if that is the case then the "kind" of energy is just "potential" rather than kinetic at the moment the object is at its highest point and it is it is very very brief, but the rest of the time the kind of energy is kinetic. The rest of the data is irrelevant unless you are measuring the height of the arc or calculating how fast the object will return given the data. The amount of kinetic energy at maximum height is zero--I don't remember how to calculate potential energy though because it is basically stationary and will have whatever potential energy a mass 1 kg happens to potentially contain depending upon the process which is used to release the energy(including nuclear fission/fusion) so basically at the height the 1 kg mass it really depends on e=mc^2 but for Newtonian physics Einstien's theories have little use so it has 0.

3. ### jav6454 macrumors P6

Joined:
Nov 14, 2007
Location:
1 Geostationary Tower Plaza
#3
Is this a book problem or teacher made problem? If it's a book problem, can I have name of book, author and edition. (I am Engineering major so I have plenty of books and notes that can help) Oh btw, chapter and problem if possible...

Joined:
Aug 15, 2006
#4
Its a teacher made problem for us to 'practice'. Would you be willing to help me set this up? I am pretty sure a diff eq is necessary, my last attempt got my a system of differential equations, something no one my age wants to deal with...

Yeah, I am looking for total energy, so the potential energy in the y direction at the apex, and the kinetic energy in the x direction. This retarding force is causing all sorts of problems for me.
Thanks, especially if someone manages to show me how to do with between now and tomorrow morning...

Also, just to clarify the retarding force is in both directions, its value will be different because the components in both directions are different though.

5. ### MorphingDragon macrumors 603

Joined:
Mar 27, 2009
Location:
The World Inbetween
#5
Gah, old Skool notation. I'm just gonna use the notation I know.

isnt Emax = Egravitypotential + Ekinetic? (Horizontal + Vertical) So you couldnt you just use RightAngle vector triangles?

This might be stupid/high school ed ignorance but, maybe you need to remind yourself of the laws of conservation. So Esigma at point one will be Esigma at point two. Draw the situation as a picture, trust me it helps.

Throwing this out there, what if Fmax = m.a - 0.12? Theres something were missing I just know it, or you're over complicating it.

6. ### CarlisleUnited macrumors 6502

Joined:
Jan 30, 2007
Location:
Nederland
#6
I don't know if you have already solved this already but could you not use the average velocity of the object in the y direction (180sin42 / 2) to figure out the average deceleration caused by the drag Fdrag = Accdrag = -0.012*120.44

Then solve for height by using
V^2 = U^2 + 2AS

Where:
V = Current velocity (0 because the object is at max height)
U = Original velocity (120.something (resolved in the y direction))
A = Acceleration, combining gravity and Accdrag (-9.81 - Accdrag)
S = Vertical displacement

Then using the value for S we can find the energy. We can ignore kinetic energy so:

E = mass x gravity x vertical displacement.

I'm not sure if that is what you were after or not.

edit: forgot about the x direction... working that bad boy out right now

7. ### crackpip macrumors regular

Joined:
Jul 23, 2002
#7
I'll give you a few pointers. I got an answer of about 6800 J, which seems reasonable given an initial kinetic energy of about 16000 J, and if there was no drag, the kinetic energy would be about 13000 J. The solution also reduces to the frictionless form if I set the drag coefficient to zero, so I think the solution is correct.

The first thing to realize is that the frictional drag does (negative) work on the projectile, i.e. it reduces the total mechanical energy. Thus, the sum of the potential and kinetic energies at the top of the arc will be less than the kinetic energy at launch.

The simplest strategy is to find the time it takes for the projectile to reach the top of its arc by integrating the y-component of the 2nd Law. Then integrate the x-component of the 2nd Law to find the x-velocity at that time.

1) Try looking for a substitution to perform the integral in the y-component of the 2nd Law.
2) Don't forget the constants of integration.
3) Make sure gravity and the drag force are acting in the proper direction.

Good Luck,
crackpip

8. ### al2o3cr macrumors regular

Joined:
Oct 14, 2009
#8
Frack - wrote up a long reply and the forum ate it.

Long and short of it, you'll just need to set up the differential equations and turn the crank. The y-axis is the trickiest, and here's a result to keep you on track:

y(t) = y_0 - (g/b)*t + (1-exp(-b*t))*(v_y0/b + g/b^2)

Here b = k/m, where F_drag = -k*v.

Surprisingly, this *does* simplify to y(t) = y_0 + v_y0*t - 0.5*g*t^2 as b->0, despite appearances.

The x equation is similar, but less messy as the nonhomogenous term there is a constant.

Hope this helps!

Joined:
Aug 15, 2006
#9
I've been working on this, and have not had any luck. Could you type, or could someone else, type out a step by step? Id really appreciate it.

I worked this through, and am getting something quite a bit larger then the poster quoted above. THe problem with this apprach is the velocity will never reach the max, so, Ill need some form of integral to take that into account.

Where is F_Drag coming in? Did you mean to include it in the above equation?

If anyone could continue with one of the posts above that would be really appreciated.

10. ### Ttownbeast macrumors 65816

Joined:
May 10, 2009
#10
I find this topic fascinating it brings me back to my first year in college....good times......good times

11. ### crackpip macrumors regular

Joined:
Jul 23, 2002
#11
Start with the 2nd Law in the y-direction (I'll use "_" to mean the next character is a subscript):
m d(v_y)/dt = -mg - b v_y => d(v_y) = -b/m * (v_y + mg/b) * dt
where b is the drag coefficient.

Next, because m, g, and b are constants, I can do the following without changing the value of the equation
d(v_y + mg/b) = -b/m * (v_y + mg/b) * dt

I can do a substitution (u = v_y + mg/b), which gives me
du/u = -b/m * dt

Integrate both sides, which will give you an undetermined constant of integration. Use the initial condition to find that constant-- at t=0 then v_y = v_0 sin(theta), where v_0 = 180 m/s and theta = 42°. Now you have the y-velocity as a function of time. Calculate the time when v_y = 0, that will be at the top of the arc.

Now do a similar calculation for the x-component of the 2nd Law and when you've got v_x as a function of time, substitute in the time calculated above. This will give you the total velocity at the top of the arc, which can be used to find the kinetic energy.

crackpip