Calculating Time of Force Application in Non-Uniform Motion

  • Thread starter Thread starter neoking77
  • Start date Start date
  • Tags Tags
    Impulse Momentum
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
neoking77
Messages
30
Reaction score
0
A 230N force causes a 4.0kg to accelerate from 5.0m/s North to 3.0m/s east. for how much time does hte force act?

t = m(vf-vi) / F

i got this but the trouble is vf and vi are in different directions...and since t is not a vector i can't break it down to components. what should i do?
 
Physics news on Phys.org
Maybe you can do it by this way.
let [tex]\vec{v_i} =(0,5)[/tex] [tex]\vec{v_f}=(3,0)[/tex]
so [tex]\Delta \vec{v} =(3,-5)[/tex]
 
You can integrate

[tex]\frac{d\vec{v}}{dt} = \frac{\vec{F}}{m}[/tex]

Your solution will be two equations (one for each component of velocity). You know the initial and final velocity vectors. Your unknowns are the two components of [itex]\vec F[/itex] and t but you also know

[tex]F^2 = F_x^2 + F_y^2[/tex]

where F = 230 N. You should be able to handle the details.
 
Last edited:
using
[tex]F_x\Delta t\ =\ m\Delta v_x\ =\ a[/tex]
[tex]F_y\Delta t\ =\ m\Delta v_y\ =\ b[/tex]
and then
[tex]a^2\ +\ b^2\ =\ \Delta t^2 \mid F\mid^2[/tex]