1its obvious yar
a force P act on body of mass 'm' intially at rest and intiates its vertical motion at t=0 .the force decreases at uniform rate and is proportional to the vertical displacement of body.Find the velocity of the body at a height 'h' from its intial postiton if 'x' is the decrease in force for unit displacement .??
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8 Answers
1dFdY=(-)x
dFdt=(-)xdYdt
F=ma
mdadt=(-)xv
dadt=(-)xmv
d2vdt2=(-)w2v.......... so v can be rewritten as::
v=Asin(wt) ..where w=√(x/m)
at t=0..a=F/m
a=Awcos(wt)........at t=0
A=a/w
so...A=F/m√m/x
y=-(Aw)cos(wt)... so .. cos(wt)=(-)yAw
v=A√1-cos2wt...using value of cos(wt)..
v=1/w√A2w2-y2
1just a doubt u said vertical motion right.........
so is gravity acting on the particle??????
1prateek even if i subsitute..w=√x/m and y=h..thn also tera anwere matchnahi ho raha
23P = - ky + c (j) ( x=k)
now WP +Wmg= change in KE
so cy - ky2/2 -mgh =1/2mv2
let P =Po at y= 0
so
c=Po
so v=\sqrt{\frac{2P_{o}y-ky^{2}-2mgh}{m}}?????????????
1thx yar...i made a mistake in getting my solution.i was getting v=[{2py-ky-mgy}/m]1/2