Deceleration due to Frictional Force

[Esran]

Homework Statement

A 1000 kg boat is traveling at 90 km/h when its engine is shut off. The magnitude of the frictional force f_k between the boat and the water is proportional to the speed v of the boat: f_k = 70v, where v is in meters per second and f_k is in Newtons. Find the time required for the boat to slow to 45 km/h.

Homework Equations

F = ma, f_k = 70v, f_k = $$\mu$$_kF_N, and various kinematic equations.

The Attempt at a Solution

First, I calculated a₀ and a_f by substituting v₀ and v_f into f_k = ma = 70v, respectively. I then calculated the average acceleration by adding the initial and final accelerations together and dividing by two. Finally, I solved the equation v_f = v₀ + a_avgt for t. The answer I got was 9.5 s, which seemed reasonable to me.

However, the back of my textbook has 9.9 s for the answer. I have a suspicion that there is a more accurate way to go about treating the acceleration than by just finding an average term. After playing around with equations for over an hour though, I'm stumped.

What is the correct method for solving this problem?

 

[turin]

Do you know differential equations? Do you know logarithms?

 

[Esran]

I know neither...or at least, I am not supposed to for this problem. ;-)

 

[Doc Al]

I have a suspicion that there is a more accurate way to go about treating the acceleration than by just finding an average term.

You'll need to set up and solve a simple differential equation, using Newton's 2nd law. Hint: a = dv/dt.

[oops... turin beat me too it!]

 

[Esran]

We're supposedly not allowed to use differential equations, as I stated before. Or logarithms.

 

[Esran]

Using differential equations would be very, very easy.

f_k = 70v
ma = 70v
a = (70v)/m
dv/dt = (70v)/m
dv/(70v) = dt/m
ln(v)/70 = 1/m
etc...

Am I correct?

 

[RoyalCat]

That's not how you set up a differential equation.
The differential equation you have here is one of the form:
$$\ddot x = -k\dot x$$

What is the solution to this equation? Once you find that, finding the velocity at a given moment should become simple.

 

[Esran]

Ah, well then, I don't know how to do it on top of not being allowed to do it.

I am still waiting for assistance on a non-differential equation approach.

 

[turin]

Using differential equations would be very, very easy.

f_k = 70v
ma = 70v
a = (70v)/m
dv/dt = (70v)/m
dv/(70v) = dt/m
ln(v)/70 = 1/m
etc...

Am I correct?

Not exactly. You're missing a minus sign. Can you tell me why?

About not being allowed to use diff. eq.:
That makes no sense. The only thing that I can think of is that you are allowed to "integrate" the solution from a first order ordinary differential form, and maybe your teacher is distinguishing this simple differential form from a differential equation in general. I would ask your teacher if you are supposed to (allowed to) use integral calculus for this problem.

There is an alternative, but I even hate to mention it, because it perpetuates mindless business-calculus approach to physics problems: maybe there is a kinematical equation in your book/notes that deals specifically with a drag force that is proportional to velocity. If that is the case, it is certainly "nonstandard". It should either have a logarithm or exponential, depending on how the equation is posed.

 

[Esran]

Okay, I'm not getting anything remotely close to 9.9 s with my differential equations. Since this problem isn't being graded, can you please show me how it's done?

 

[Doc Al]

Okay, I'm not getting anything remotely close to 9.9 s with my differential equations. Since this problem isn't being graded, can you please show me how it's done?

Did you correct your errors in post #6? You were on the right track, but had a couple of errors: a missing minus sign and an improper integration on the right hand side (probably a typo).

 

[Esran]

Yes I did. Here are my calculations:

f_k = 70v
m(-a) = 70v
-a = (70v)/m
-dv/dt = (70v)/m
-dv/(70v) = dt/m
-ln(v)/70 = t/m

t = m(-ln(v)/70)

$$\Delta$$t = 1000(-ln(12.5)/70) - 1000(-ln(25)/70)
$$\Delta$$t = 9.9 s

Hehe, oooops, so I did get it. I must have been entering some numbers wrong.

 

[Doc Al]

Sweet.

 

[Esran]

Alright, even though I got the answer right, if I'm understanding correctly, there's supposedly a way to do this problem just by fidgeting around with various equations. I haven't the slightest clue how...maybe set up a pair of equations and cancel out a variable?

 

[turin]

Formally, it doesn't matter where you put the minus sign, but I just want to comment that, physically, I would argue that the minus sign should go on the RHS. The equation is basically Newton's 2nd law, and the RHS is the total applied force (in the plane of the water). So, the minus sign on the RHS would indicate that the applied force is in the opposite direction to the velocity. I suppose you probably reasoned that the boat decelerates, so the acceleration is negative, and that works for 1-D, but this kind of distinction doesn't work in more than 1-D, so be careful.

Also, I'm still quite curious, how are you supposed to do this without diff. eq.?

edit: Oh, I see that Esran is also still curious. Esran, what book are you using? From where did you get this problem? Maybe, there is a formula for exponential decay that you are supposed to know applies to this problem.

For example, I don't think radiological technicians are "supposed to" know diff. eq., but they "know" that activity decays exponentially over time. And freshman aren't "supposed to" know diff. eq., but they are supposed to know that the voltage and current in an RC circuit follow exponential curves.

 

[Esran]

I don't know, I'm about to phone my old physics teacher from high school to see if she knows how.

 

[Doc Al]

Formally, it doesn't matter where you put the minus sign, but I just want to comment that, physically, I would argue that the minus sign should go on the RHS.

Good point. I agree.

 

[RoyalCat]

Alright, even though I got the answer right, if I'm understanding correctly, there's supposedly a way to do this problem just by fidgeting around with various equations. I haven't the slightest clue how...maybe set up a pair of equations and cancel out a variable?

Well, solving a differential equation is pretty much that. You're asking yourself, what function can I think of that if I differentiate it once, I get it back but multiplied by a negative constant.

You then fidget around with functions you think would do the trick until you find it.

As has already been posted (In one way or another) and since the problem has been solved:
$$\ddot x = -k\dot x$$

We'll assume that the answer is of the form:

$$\dot x(t)=Ae^{-kt}$$

Let's check if we hit the jackpot (Note that the initial conditions could mean that my answer, though it is correct for DE, might just be a private case, for instance, 0, which would mean I need to 'guess' a better function):
$$\ddot x = -kAe^{-kt}=-k\dot x$$

Now to find A:
$$\dot x(0)=90 km/h$$
$$A=90 km/h$$

$$k=70/1000 Hz$$

Now we have a complete description of the behavior of the boat (Almost, integrating this result and looking at the initial conditions for x(t) to find the integration constant would give us the real complete description):
$$\dot x(t)=v(t)=90*e^{\tfrac{7}{100}t}$$ $$\frac{km}{h}$$

Plugging in 45 km/h would net you the exact same answer you got by integrating.

 

[turin]

Well, solving a differential equation is pretty much that. You're asking yourself, what function can I think of that if I differentiate it once, I get it back but multiplied by a negative constant.

You then fidget around with functions you think would do the trick until you find it.

Not with a first order ordinary diff. eq., which is the case in this problem. That has a formal solution without guessing (or "fidgetting around").

 

[kmohan24]

I have a variation of this problem, this is from an entrance exam in India where you need to do this problem in a minute.
---
A particle of mass m is at x = 0 and moving along the x-axis with velocity v0, at time t = 0. It is subjected to a frictional force –bvx where b is a constant and vx is the velocity in the x-direction. At what position x on the x-axis will it come to rest?

a. x = bv0
b. x = (bv0)/m
c. x = mbv0
d. x = (mv0)/b
---

Can this be done without Diff. Equations? This question is at the high school level. Intitutively I thought (a) is the answer, since this is not dependent on the mass of the particle (even though a = f/m).

What is best method to do this problem?

Krish Mohan

---

Homework Statement

A 1000 kg boat is traveling at 90 km/h when its engine is shut off. The magnitude of the frictional force f_k between the boat and the water is proportional to the speed v of the boat: f_k = 70v, where v is in meters per second and f_k is in Newtons. Find the time required for the boat to slow to 45 km/h.

Homework Equations

F = ma, f_k = 70v, f_k = $$\mu$$_kF_N, and various kinematic equations.

The Attempt at a Solution

First, I calculated a₀ and a_f by substituting v₀ and v_f into f_k = ma = 70v, respectively. I then calculated the average acceleration by adding the initial and final accelerations together and dividing by two. Finally, I solved the equation v_f = v₀ + a_avgt for t. The answer I got was 9.5 s, which seemed reasonable to me.

However, the back of my textbook has 9.9 s for the answer. I have a suspicion that there is a more accurate way to go about treating the acceleration than by just finding an average term. After playing around with equations for over an hour though, I'm stumped.

What is the correct method for solving this problem?

 

[RoyalCat]

You've got two options for that sort of exercise.

The first is dimensional analysis. This is a textbook example of where dimensional analysis comes into play.

The units of b are:

$$F=-bV$$
[tex]=[F][V]^{-1}[/tex]
[tex]=[M][T][/tex]

So see if you can use dimensional analysis to solve it from here on out.

The second is just integration. This is a very simple exercise in that as well.

Writing out NSL for that particle:

$$ma=-bv$$

$$k\equiv \tfrac{b}{m}$$

$$\ddot x = -k\dot x$$

$$\frac{dv}{dt}=-k\frac{dx}{dt}$$

And from here you can probably see $$v(x)$$ straight away.

And to turin:
Thanks for that comment, using integration instead of guessing is so much simpler most of the time.