# a particle of mass (1/100) kg is moving along the positive x-axis under the influence of a force F(x)= (-K/2x^2), where k=1/100 Nm^2.AT time t=0 it is at x=1m and it’s velocity is v=0;(A) Find it’s velocity when it reaches x=0.50m?(B) Find the time at which it reaches x=0.25m?

Vasantha Kumari
8 years ago

Arun Kumar IIT Delhi
8 years ago
Hi
by what i understand by your representation of equation.
$\\F(x)=-\frac{kx^2}{2} \\a=-x^2/2 \\vdv/dx=-x^2/2 \\=>v^2=-x^3/3+c \\=>0=-1/3+c \\=>c=1/3 \\=>v^2=-\frac{x^3}{3}+\frac{1}{3} \\=>v^2_{at\,x=0.5}=-\frac{1}{8*3}+\frac{1}{3} \\=>v^2_{at\,x=0.5}=\frac{7}{8*3} \\=>v_{at\,x=0.5}=\sqrt{\frac{7}{24}}$
$\\\frac{dx}{dt}=\sqrt{\frac{-x^3}{3}+\frac{1}{3}} \\\frac{dx}{\sqrt{\frac{-x^3}{3}+\frac{1}{3}}}=dt$
It would be great if you can integrate it with JEE methods.
Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty
Arun Kumar IIT Delhi
$\\F(x)=-\frac{k}{2x^2} \\=>a=-\frac{k}{2x^2} \\=>vdv/dx=-\frac{k}{2x^2} \\=>\frac{v^2}{2}=\frac{1}{2x}+c \\=>0=\frac{1}{2}+c \\=>c=-1/2 \\=>\frac{v^2}{2}=\frac{1}{2x}-\frac{1}{2} \\=>v^2=\frac{1}{x}-1$
$\\=>v^2=\frac{1}{x}-1 \\=>\frac{dx}{dt}=\sqrt{\frac{1}{x}-1} \\=>\frac{dx}{\sqrt{\frac{1}{x}-1}}=dt \\=>1/x-1>0 \\=>1/x>1 \\=>x<1 \\lets assume x=sin^2\phi$
$\\=>sin2\phi d\phi =dx \\=>tan\phi *sin2\phi d\phi =\frac{1-cos2\phi}{2}d\phi \\=>integrating \\=>\frac{\phi-sin2\phi/2}{2}$