1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A truck with growing mass

  1. Jun 14, 2006 #1
    Ok, I think I might have missed a key point in this class from a few weeks ago but anyways.

    There is a truck of mass M, who has a constant force F applied to it. The truck is also being filled with coal at a constant rate b kg/s. I'm supposed to find an expression for the velocity of the truck.

    First I tried:

    F = ma

    a = F/m

    Then try to integrate both sides w.r.t time to get velocity, but since the mass depends on time as well, I get an expression involving ln(m).

    So then I tried it this way:

    F = dp/dt

    Integrate both sides,

    Ft = p = mv = (M+bt)v

    v = Ft/(M+bt) which is the desired result.

    I think I remember something from class about the Force more accurately being the change in momentum, and not necessarily the mass-acceleration product. Was this question just to emphasis that point? I think I'll have to go read the text :confused:
  2. jcsd
  3. Jun 14, 2006 #2


    User Avatar
    Gold Member

    Newton's second law says [tex]\vec{F}_{net}=\frac{d\vec{p}}{dt}[/tex]

    Since [tex]p=mv[/tex] you have [tex]\vec{F}_{net}=\frac{d}{dt}(m\vec{v})[/tex]

    If you apply the product rule you obtain [tex]\vec{F}_{net}=\vec{v}\frac{dm}{dt} \ + \ m\frac{d\vec{v}}{dt}[/tex]

    If the mass is constant (ie dm/dt=0) then you are left with [tex]\vec{F}_{net}=m\frac{d\vec{v}}{dt}[/tex]. Of course [tex]\frac{d\vec{v}}{dt}[/tex] is just acceleration so you are left with [tex]\vec{F}_{net}=m\vec{a}[/tex]

    Edit: Also, is the answer you obtained given as the correct answer in the book?

    If you consider [tex]\vec{F}_{net}=\vec{v}\frac{dm}{dt} \ + \ m\frac{d\vec{v}}{dt}[/tex] it looks like you would get a different answer.
    Last edited: Jun 14, 2006
  4. Jun 14, 2006 #3
    Hmm.. now that I reread the question, it's asking to show that the speed is Ft/(M+bt)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?