Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

A partical is projected along a horizontal field whose co-efficient of friction varies $\mu = \large\frac{A}{x^2}$ where x is a distance from origin in meters and A is a positive constant. The initial distance of particle is 1m from orgin and its velocity is radially outwards. The minumum initial velocity at this point so that the particle never stops is

\[(a)\;\sqrt {2gA} \quad (b)\;\infty \quad (c)\; 2 \sqrt {gA} \quad (d)\;4 \sqrt {gA}\]
Download clay6 mobile app

1 Answer

The body must start from $x=1$.Then, the work done against friction must be equal to the initial kinetic energy given to the body in the form of initial velocity of the body
$\large\frac{1}{2}$$mv^2=\int \limits_1^{\infty} \mu mg \;dx$
$\large\frac{v^2}{2}$$=Ag \int \limits_1^{\infty} \large\frac{1}{x^2}$$dx$
$\large\frac{v^2}{2}$$=Ag \bigg[ \large\frac{-1}{x}\bigg] _1 ^{\infty}$
$v^2=2 Ag$
$v=\sqrt {2gA}$
Hence a is the correct answer.
answered Aug 7, 2013 by meena.p
edited Jun 13, 2014 by lmohan717

Related questions