(simplified?) Air drag force question.

[dizweld]

My issue/knowledge
Hello all, I think this is my first time posting a question here (I have read the guidelines). This problem is not in my textbook or notes, and I'm not sure how to go about it. Online all I can find about air drag equations requires knowing variables which I'm not given here, so I'm having a great deal of difficulty (for 6 hrs+ now). I'm sorry if my attempts look weak too you, I have really tried. Thanks.

Homework Statement

"A 1500kg car accelerates in first gear to from 0 to 55km/h in 2.7 seconds. In fifth gear from 150 km/h to 220 km/h in 23 seconds. Assume the drag force is negligible at low speeds but increases as the square of the velocity. Estimate the drag force of friction at a speed of ~200 km/h"

Homework Equations

v^2?
F=m*a
a=F/m
a=v/t
fk=ukmg
uk= a/g

The Attempt at a Solution

a=v/t
(55km/h-0km/h) / 2.7 seconds= 5.658 m/s^2 (converted to metric)
(220km/h-150km/h) / 23 seconds= 0.845 m/s^2
m*a
1500kg*5.658m/s^2= 8487N
1500kg*0.845 m/s^2= 1267.5N

So I get here and have no idea how to proceed. What I feel like maybe I should do is imagine it like a normal friction problem and just find acceleration it takes to 200km/h, divide by gravity, uk=a/g. Then again I'm thinking that square of velocity is really important, so maybe subtract force of car it takes to get to 200km/h from 200km/h^2, then divide by mass to get acceleration, divide that by gravity? I feel like I'm talking nonesense now.

 

[voko]

Gravity plays no role here.

What does play a role is the force on the car due to its engine, and the force of drag. How are they related?

 

[dizweld]

Gravity plays no role here.

What does play a role is the force on the car due to its engine, and the force of drag. How are they related?

Well force of drag goes opposite direction of force of car, so in order to accelerate to higher velocities more force car is required to achieve that acceleration? Hm

 

[voko]

What is the relationship among the force due to the engine, the drag, and acceleration?

 

[dizweld]

Has to be mass right? Maybe velocity is a better answer.

 

[voko]

"Relationship" means "equation" or "equations"; "mass" is a number.

 

[dizweld]

Oh sorry. Well a=f/m , and both force due to engine and drag are forces so f=a*m

 

[dizweld]

What about this?

So if square of velocity is the force of drag, then (200km/h)^2--->(55.55m/s)^2= 3086.4N
Then since apparently engine force is 1267.5N, 1267.5N-3086.4N= -1818.9N

I know I'm probably wrong...If I divided by mass that would give acceleration but I don't think that is drag force of friction. *sigh*

 

[voko]

What makes you think that the engine force is 1267.5 N?

Note the problem said "Assume the drag force is negligible at low speeds".

 

[dizweld]

Assuming 0-55km/h is low speed and 150 to 220km/h is high speed, so I calculated the acceleration from 150 to 220 multiplied it by 1500kg after conversion and got that force.

edit: 0-55 in 2.7 seconds is a quick acceleration hinting that there is negligible drag force...The only data left is 150 to 220 in 23 seconds which is slow hinting drag force.

 

[voko]

The drag force is NOT negligible at high speeds. 1267.5 N is the sum of the engine and drag forces.

 

[dizweld]

I see that now wow. I was, still am confused by "increases by square of velocity".

 

[voko]

So, what is the force of the engine at the low speed?

 

[dizweld]

8487n.

 

[voko]

Assuming the engine produces the same power at the high speed, what is the engine force at the high speed?

 

[dizweld]

Tried employing work energy theorem but I'm not there yet so...trying to figure out how to do it another way, difficult for me heh.

 

[voko]

Might be useful: Power = Force x Velocity.

 

[dizweld]

So alright here is what I'm doing. 8487N*(15.277m/s)=129662.5P

If the power is the same at a velocity of 220km/h->61.11m/s
Then I can do F=P/velocity, but that gives me =2121.788N? Which is higher than 1267.5N the sum of engine and drag. Is that to say engine power is 2121N and drag is the subtraction result which is -854.288?

Feel like I'm getting something seriously wrong but I never did these equations before.

 

[voko]

It seems to me that this problem is very badly formulated.

Using the work-energy theorem.

The initial velocity is zero and the final velocity is 55 km/h, hence the change in kinetic energy is 175 kJ, divided by 2.7 s, that gives 65 kW of power.

At the high speed, the initial speed is 150 km/h, the final is 220 km/h, the energy diff is 1.5 MJ. The work of the motive force is 65 kW over 23 s, yielding 1.5 MJ, too. This suggests that the work of the force of drag is zero, which means that there is no drag at the high speed, which obviously is nonsensical. Please check that the numbers you have are copied correctly.

 

[dizweld]

"A 1500kg car accelerates in first gear to from 0 to 55km/hour in 2.7 seconds. In 5th gear the car accelerates from 150km/h to 220km/h in 23 seconds. Assume the drag force from wind is negligible at low speeds but increases as the square of the velocity. Assuming there is negligible wind drag in 1st gear, estimate the drag force of friction at a speed of ~200km/h?"

Been staring at this thing for so long now, but I recopied just now. Only difference is that I left "Assuming there is negligible wind drag in 1st gear" out because it seemed rhetorical. Otherwise yeah everything is exact.

 

[dizweld]

My prof. has stated it's possible to use kinematics/motion for this? If that helps at all. Yeah this problem is very frustrating to me.

 

[voko]

I am really unsure what other assumptions you should be making here. The point of this problem, as it seems to me, is to find out the motive force or power at the low speed, and then somehow figure out what the motive force or power should be at the high speed, and with that, find the drag.

The assumption that seemed natural to me was to assume that the power is same at low and high speeds. But that assumption results in the conclusion that there is no drag at the high speed. I have no other ideas. Perhaps you could discuss that with your professor. Maybe someone else here could comment on this.

 

[ehild]

Try to assume constant force for the engine.

ehild

 

[dizweld]

I have with no luck unfortunately, not sure how to use that.

 

[ehild]

You have calculated the force that accelerated the car from zero to 55 km/hour in 2.7 s: it was
8487 N. You also determined the force that accelerated the car from 150 to 220 km/h in 23 s, it was ma=1267.5N.

At low speed the whole force accelerates the car. If the force exerted by the engine is constant (it is torque exerted really) F(engine) = 8487 N. At high speed, the drag is opposite to the driving force, so ma = F(engine)-F(drag) . What is the drag force then?

ehild

 

[dizweld]

I tried this once rearranging equation to -Fdrag = ma - Fengine= 1264.5N-8487N= -7222.5N

It seemed too much to me, so maybe I am not rearranging properly. Cant see another way to do it, can it be really that simple an answer?

 

[ehild]

I tried this once rearranging equation to -Fdrag = ma - Fengine= 1264.5N-8487N= -7222.5N

It seemed too much to me, so maybe I am not rearranging properly. Cant see another way to do it, can it be really that simple an answer?

You said :

-Fdrag = -7222.5N

What is Fdrag then ?

You might write and solve a differential equation for the high-speed case, but the problem asks you to estimate the drag force. I think, it is that simple. If not, we can try that differential equation.

ehild

 

[dizweld]

Thanks to the both of you for the help. I will find out what he expected once it is graded I guess ^ ^.