m1 dynamics fast question

[7ooD]

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

[Amr Fouad]

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 

[7ooD]

ya gammed thx ya basha

[Amr Fouad]

any time