Motion with frictional force but without driving force

  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. kuruman

    kuruman 3,448
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    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. 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. kuruman

    kuruman 3,448
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    Forget μk. Write down Newton's Second Law for the motion of this mass.
     
  6. 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. kuruman

    kuruman 3,448
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    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. 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. kuruman

    kuruman 3,448
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    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. 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. vela

    vela 12,444
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    That's not correct since you've made some really basic algebra errors. Try again.
     
  12. kuruman

    kuruman 3,448
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    Incorrect. How did you get the second line from the first?
     
  13. 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. kuruman

    kuruman 3,448
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    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. 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. kuruman

    kuruman 3,448
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    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. 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. kuruman

    kuruman 3,448
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    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. (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. kuruman

    kuruman 3,448
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    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. hm.. so the acceleration is not constant.. does that means i need to derive acceleration function?
     
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