IIT JEE , AIEEE Forum - Physics , Mechanics

amitp91

a particle of mass 10^-2kg is moving along the positive x axis under the influence of a force F(x) = -k/2x² where k=10^-2Nm². at time T=0 it is at x=1.0m and its velocity is v=0 . its velocity when it reaches x=0.5m
a)0.5m/s along +xaxis
b)1m/salong +x axis
c)0.5m/s along - xaxis
d)1m/s along - xaxis

elastiboysai

edited

mvdv/dx=-k/2x^2

arrange n integrate

[-k/2mx](frm lim x0 to x1) = [v^2/2] frm lim v0 to v1

v0--veloc at time t=0..

substitute values

n u get

-1

mistake corr.

amitp91

i think u are wrong

anchitsaini

mvdv/dx=-k/2x^2
as m= k
hence
vdv/dx = -1/2x^2
-2vdv=dx x^-2
-v^2 = -1/x
hence
v^2 =1/x   from 1 to 0.5
      =1/0.5 -1
      =1
hence
the answer seems to be -1m/s

BALGANESH

edited

F = -1/ 200x^2

a = -1/ m200x^2

a = - 1/2x^2

(dv/dt) . dx/dx =-1/2x^2

vdv = - 1 dx /2x^2

on integrating LHS ( V1 TO V2 WHERE V1 = INITIAL VEL = 0 ) AND RHS x1 to x2 where x1 = 1 and x2 = 0.5

-(V2 )^2 =1/2 [ 1/0.5 - 1/1 ] ( minus sign has been taken into consideration )

-V2 ^2/2 = -0.5

V2 = 1M/S ALONG +X-AXIS

anchitsaini

quote from balganesh --
(V2 )^2 =1/2 [ 2/0.5 - 2/0.1 ] ( minus sign has been taken into consideration )

it should be
(V2 )^2 / 2

anchitsaini

another mistake--
(V2 )^2 =1/2 [ 2/0.5 - 2/1 ] ( minus sign has been taken into consideration )

it should be
(V2 )^2 / 2 =1/2 [ 1/0.5 - 1/1 ]

MUDIT

edited

karthik2007

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