- #1
speg
- 16
- 0
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
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