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Kinetic friction in terms of velocity

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A 1150 kg boat is traveling at 100 km/h when its engine is shut off. The magnitude of the frictional force k between boat and water is proportional to the speed v of the boat: f_k = 73v, where v is in meters per second and f_k is in newtons. Find the time required for the boat to slow to 41 km/h.

    2. Relevant equations



    3. The attempt at a solution

    Quite honestly, I don't even know where to begin with this problem. I started to write down:

    F_net,x = (1150)*(a) = F_app - f_k

    Since we are dealing with the force of friction f_k against the applied force of the boat's engin, F_app. So we would solve for 'a', but the applied force F_app is unknown, and the friction is dependent on the current velocity, so it's not exactly applicable unless we have the velocity at every given point.. Makes me think I have to take it's integral, but that would be it's position, not it's acceleration..
     
  2. jcsd
  3. Oct 2, 2009 #2
    You are dealing with the wrong Newton's Second Law.

    You are asked about what happens after the engine is shut off. What is the force acting on the boat from that point on?

    State Newton's Law differentially [tex]ma=m\frac{dv}{dt}[/tex] and do the same for the net force. Equate the two, use the chain rule as needed, and find the velocity as a function of time.

    Alternatively, if you know a better way to solve the kind of differential equation you get, you could try doing that instead.
     
  4. Oct 2, 2009 #3
    There is a hint provided that stated exactly that identity, however, now that we have time, it's not quite clear how to use dv or dt separately:

    m(dv / dt) = F_app - f_k

    Since we have turned off the engine, then F_app = 0. Also, f_k = 73v, as stated:

    m(dv / dt) = - 73v

    m*dv / (-73v) = dt

    Now, I'm assuming since dv and dt have no meaning outside of the derivative, then I'll switch to the discrete differences: dv = v_f - v_i and dt = t_f - t_i. But t_i = so then dt = t_f or just t

    (1150)*(v_f - v_i) / (-73*v) = t

    Also, it's not clear how to incorporate v, since it changes over time. But I'm certain there isn't any differential equations involved here, since this is just introductory physics. So I tried both 100 km/h = 27.7 m/s and 41 km/h = 13.8 m/s as values for v and neither worked:

    1150*(13.8 - 27.7) / (73*27.7) = t
    7.916 = t

    and

    1150*(13.8 - 27.7) / (73*13.8) = t
    15.9 = t
     
  5. Oct 3, 2009 #4
    [tex]dv[/tex] and [tex]dt[/tex] have plenty of meaning!

    You got so close, all you needed to do was integrate!

    [tex]-k\frac{dv}{V}=dt[/tex]

    [tex]k\equiv \frac{m}{b}[/tex] where [tex]m[/tex] is the mass of the boat and [tex]b[/tex] is the drag coefficient (73 in our case)
     
  6. Oct 3, 2009 #5
    Ack! Okay, my Calc II teacher really didn't cover separable differential equations as well as he could have. He especially didn't cover their applications. I'll probably have to pay for his incompetence and study it better when I have the time, but for right now, here's what I have:

    [tex]F_{net} = m*\frac{dv}{dt}[/tex], by definition. Since the motor is turned off, [tex]F_{app} = 0[/tex], and the opposing frictional force, [tex]f_k(v) = 73v[/tex]. We then have

    [tex]m*\frac{dv}{dt} = 0 - 73v[/tex]

    or

    [tex]\frac{m*dv}{-73v} = dt[/tex]

    Integrating both sides

    [tex]\frac{m}{-73} \int \frac{1}{v} dv = \int dt = \frac{m}{-73} \ln |v| + C = t[/tex]

    When t = 0, v = 27.7, so solving for C

    [tex]\frac{-1150}{73}* \ln(27.7) = -C = -52.32[/tex]

    Now we must find [tex]t = t_f[/tex] where v = 13.78.

    [tex]\frac{-1150}{73}* \ln(13.78) + 52.32 = t_f = 10.99[/tex]

    But this answer doesn't work.

    Either a) I screwed up in my calculations, or b) my lack of experience with differentials, I don't have something set up properly.
     
  7. Oct 3, 2009 #6
    Just as a rule of thumb, work parametrically. It's easy to lose your constants when you work numerically, and as a result, it's very hard for someone looking at your work to follow what you were doing. I'm afraid I got lost along the way.

    You made a mistake when you took the indefinite integral instead of a definite one (Not that it's a mistake, it just means you need to do a lot more work):

    [tex]-k\frac{dv}{V}=dt[/tex]
    For the RHS, I'll integrate from [tex]V=V_0[/tex] to [tex]V=V(t)[/tex], for the LHS, I'll integrate from [tex]t=0[/tex] to [tex]t=t[/tex]

    [tex]-k \int_{v_0}^{v(t)} \frac{dv}{V}=\int_{0}^{t} dt[/tex]

    [tex]-k \left[\ln{V}\right]_{v_0}^{v(t)}=\left[t\right]_{0}^{t}[/tex]

    [tex]-k(\ln{v(t)}-\ln{v_0})=t[/tex]

    In more compact form:

    [tex]k\cdot\ln{\frac{v_0}{v(t)}}=t[/tex]

    And from there you can get your final answer.

    Developing that expression further we can find:

    [tex]v(t)=v_0\cdot e^{-\tfrac{t}{k}}[/tex]

    I'll now show the second method I mentioned in my first post.

    Let's look at the first differential equation we got:

    [tex]\frac{dv}{dt}=-\tfrac{1}{k}v[/tex]

    Rewriting using Newton's notation we get:

    [tex]\ddot x = -\tfrac{1}{k}\dot x[/tex]

    This is a differential equation that asks the question, what function of time, when differentiated once with respect to time, gives itself times a negative constant?

    An easy solution to guess would be the exponential we got at the end of the integration above. Checking for the initial values would confirm that it is the true solution and that we haven't missed anything with our educated guess.

    You may be familiar with the procedure from how you derive SHM in class (You get the following differential equation [tex]\ddot x = -\omega^2 x[/tex] and guess the harmonic function as a solution to the diff. equation)
    Or if you're not, now you'll have a head-start once you get to that subject. :)

    Oh, and you said a little something that bugged me. You said that [tex]\Sigma \vec F = m\vec a[/tex] by definition

    The above equation is Newton's Second Law. It should not be taken for granted. :)
     
  8. Oct 3, 2009 #7
    Wow.. thank you so much!! Yes, thank you for pointing out the numerical issue. I just realized that my conversion between km/h and m/s was off on the final velocity. So it wasn't my calculus but my multiplication.. duh :/

    Believe it or not I actually attempted to use a definite integral like you describe, in my first attempt, but I assumed my method was wrong because my numbers were wrong, but looking back, the conversion was wrong there too, so I end up with the same solution if I used the proper values. So in the end, it looks like they both had the same answer.

    Oh, yea, and I'm aware that Newton's second law isn't a definition. I've just got used to saying that from writing lots of proofs..

    Suffice to say, I actually enjoy calculus a bit more than physics, but some of these word problems just ties my brain in knots.. ah well..

    Thank you again for the help!
     
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