Author Topic: m1 dynamics fast question  (Read 805 times)

Offline 7ooD

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    • 7ood
m1 dynamics fast question
« on: March 31, 2010, 02:05:29 am »
the question is divided into two parts the hardest part i solved my own the second i cnt think of

a particle of mass 2kg rests in limiting equilibrium on a plane inclined at 25 to the horizontal the angle of inclination is decreased to 20 and a force of magnitude 20N is applied up a line of greatest slope find the particle's acceleration. when the particle has been moving for 2 seconds the force is removed determine the further distance the particle will move up the plane

the coefficient of friction is 0.466307658
acceleration is 2.353981116
final answer should be 1.45m
i dnt understand the bold part plz i need it for tommorow
« Last Edit: March 31, 2010, 02:09:00 am by 7ood »
pimpin ain't dead it just moved to the web



Amr Fouad

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Re: m1 dynamics fast question
« Reply #1 on: March 31, 2010, 11:38:58 am »
OK..HER'S THE DEAL..

u've got to find the speed,v, when the force is removed..:

v=u+at
v=2.357x2= 4.714 m/s

then u have to find the new acceleration:

-mgsin20-uR=ma..
-mgsin20-umgcos20=ma   (divide by m)
-9.8sin20-0.466x9.8xcos20=-7.643 (the -ve sign is due to opp direction)

now:

v^2=u^2n+2as..

v=0
u=4.714
a=-7.643

therefore:

s= -(4.714)^2/2x-7.643

s= 1.45 m...hope that helped  ;D


Offline 7ooD

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Re: m1 dynamics fast question
« Reply #2 on: March 31, 2010, 12:05:48 pm »
ya gammed thx ya basha
pimpin ain't dead it just moved to the web



Amr Fouad

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Re: m1 dynamics fast question
« Reply #3 on: March 31, 2010, 12:12:07 pm »
any time   ;)