Motion with frictional force but without driving force

  1. Jul 12, 2010 #1
    Hi guyz... im having a problem with this question. Hope to get some guideline from here :D

    An engine of mass m moves without driving force but under the influence of the frictional force f(v)=(alpha)+(Beta)v2 on horizontal rails. Let the initial velocity be v0.

    (a) After which time does the engine come to rest? What is the maximum deceleration time (v0 approaching infinity )

    (b)What distance has then been covered?
     
    Last edited: Jul 12, 2010
  2. jcsd
  3. Jul 12, 2010 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Hi LockeZz and welcome to PF. Please follow the rules of this forum and use the template when you seek help with homework. Show us the relevant equations and tell us what you tried and what you think about the problem. We just don't give answers away.
     
  4. Jul 12, 2010 #3
    Sorry... for being short in giving info..

    given the frictional force function is f(v)=a+Bv2 .

    i tried to substitute the initial velocity in side the function:

    f(v0)=a+Bv20
    ukFn=a+Bv20.

    im stucked at here.. and i was thinking that i might need to get an expression which include the time..
     
    Last edited: Jul 12, 2010
  5. Jul 12, 2010 #4

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Forget μk. Write down Newton's Second Law for the motion of this mass.
     
  6. Jul 12, 2010 #5
    erm.. i want to ask about this.. how should i type in the greek symbol since the expression given a+bv^2 are actually represent by alpha and beta .

    ----------------------------------------------------------------------------------------

    so.. newton 2nd law given F=ma .
    should i substitute into the expression to get :

    ma=(alpha)+(Beta)v20

    ?
     
  7. Jul 12, 2010 #6

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    You need to write an expression that is true for all time, not just t = 0. The velocity is v0 only initially. Also you need to express the acceleration in terms of the velocity. What is the appropriate definition?
     
  8. Jul 12, 2010 #7
    Sorry for being messing up just now..For i have already edited the 1st post..

    --------------------------------------------------------------------------------------------------

    for the general expression for all time, i obtain this :

    (alpha)+(beta)v2=ma

    a=[((alpha)+(beta)v2)/m]

    so, i have try on further which i substitute the final velocity (which is 0) for part (a) and i get :

    a=(alpha)/m
     
  9. Jul 12, 2010 #8

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    OK, but you did not write down the definition of the acceleration. That is a = dv/dt. So you get a differential equation

    [tex]m\frac{dv}{dt}=\alpha+\beta v^2[/tex]

    Can you solve this to get v(t), that is the velocity for any time t?
     
  10. Jul 13, 2010 #9
    alright, for the v(t) function i derive from:

    [tex]
    m\frac{dv}{dt}=\alpha+\beta v^2
    [/tex]

    [tex]
    \int\frac{dv}{v^2}=\int(\frac{\alpha+\beta}{m})dt
    [/tex]

    [tex]
    v=\frac{-m}{(\alpha+\beta)t}
    [/tex]
     
  11. Jul 13, 2010 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    That's not correct since you've made some really basic algebra errors. Try again.
     
  12. Jul 13, 2010 #11

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Incorrect. How did you get the second line from the first?
     
  13. Jul 13, 2010 #12
    Haha.. what a silly mistake i have done.. Well i have try again on the derivation and i get this :

    [tex]
    \int \frac {m}{\alpha+\beta v^2} dv =\int dt
    [/tex]

    and im need help on integrate this.. :cry:
     
  14. Jul 13, 2010 #13

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Move the mass m to the other side. Then do the integral by substitution. Try

    [tex]v=\sqrt{\frac{\alpha}{\beta}} \ tan \theta[/tex]
     
  15. Jul 14, 2010 #14
    Thx for reminding.. i have substitute with tangent theta and get the expression u shown.. then i do the further integration and i get this :

    [tex]

    v=\alpha(\frac{t}{m}+c)
    [/tex]
     
  16. Jul 14, 2010 #15

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Check your math. Your expression says that the speed varies linearly with time. That's the case only when the acceleration is constant. Is the acceleration constant in this example?
     
  17. Jul 14, 2010 #16
    i'm not quite sure in that case.. the 1st post shows all the statement given on the example.. and its tricky and confuse for me :(
     
  18. Jul 14, 2010 #17

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    There are no tricks involved, just organized thinking. Answer these questions

    1. Does the velocity change? (Yes/No)
    2. Does the force change? (Yes/No)
    3. What is an expression for the acceleration in terms of the force?
    4. Assuming that you correctly answered all of the above, is the acceleration changing? (Yes/No)
     
  19. Jul 14, 2010 #18

    (1) yes the velocity is changing.
    (2) the force is changing since the velocity is involve in the function of force
    (3) F=ma
    (4) it should be constant (i think) since it have stated the initial velocity to be v0 and then the engine stop its motion..
     
  20. Jul 14, 2010 #19

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Correct.
    Correct.
    Correct.
    Incorrect. The initial velocity is v0, not the velocity forever and ever. As you stated in (1) the velocity changes, therefore it cannot stay at v0. Now look at (2) and (3)

    a = F/m. If F changes and m stays the same, can the acceleration stay the same?
     
  21. Jul 14, 2010 #20
    hm.. so the acceleration is not constant.. does that means i need to derive acceleration function?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Motion with frictional force but without driving force
Loading...