Posted by rav on Saturday, July 15, 2006 at 1:05pm.
A particle of mass m kg is acted on by two forces F1 and F2 with magnitudes 3root5 newtons and root5 newtons ad drections parallel to the vectors i+2j and i-2j respectively
The particle is initially at a position given by vector 2i+j
iv calculated the cartesian components of F1 and F2
F1 = 3i+6j
F2 = i 2j
iv also calculated the cartesian component of the total force F1+F2 = (3i+6j)+(i-2j) = 4i +4j
however the part im struggling on is that now the particle is of mass 1kg and is initially at rest, i have to use newtons 2nd law to write down an equation of motion and find the position of the particle after 1second.
so far iv tried:
F=ma
F1+F2= 1a
therefore a = 4i+4j but then i dont know how to find the new posistion ...can anyonehelp thankz
New postion= oldposition +intialvelocity*time + 1/2 acceleration*time^2. Technically, you should integrate to get this equation, but because acceleration is constant, the integration leads to the preceeding.
where acceleration= force/mass, and you have force and mass.
all the above of course is a vector equation, not messy at all. You can work the i,j components as separate equations, as the system is orthogonal. If needed, I can critique your work.
thankz iv worked it out
Assuming you have calculated the "a" vector correctly, just multiply it by t^2 to get the vector position change at time t. You do not have to add a term for (initial velocty vector) times t,, because the mass was initially at rest.
Add the position change vector to the initial position coordinates to get the new location.
I forgot about the factor (1/2) when calculating the position change due to acceleration. Bob Pursley's answer, posted just before mine, is correct
I have no clue.
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