What is the expression for the velocity of a truck with growing mass?

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The discussion focuses on deriving the velocity expression for a truck with a growing mass due to a constant force and a constant rate of coal being added. Initially, the approach using F = ma leads to complications because mass is not constant. The correct method involves using the momentum principle, where the net force is defined as the rate of change of momentum. By integrating the force over time and applying the product rule, the final expression for velocity is derived as v = Ft/(M+bt). The conversation emphasizes understanding the relationship between force, mass, and momentum in dynamic systems.
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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:
 
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Newton's second law says \vec{F}_{net}=\frac{d\vec{p}}{dt}

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

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

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

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

If you consider \vec{F}_{net}=\vec{v}\frac{dm}{dt} \ + \ m\frac{d\vec{v}}{dt} it looks like you would get a different answer.
 
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Hmm.. now that I reread the question, it's asking to show that the speed is Ft/(M+bt)
 
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