Question

A mass m sliding horizontally is subject to a viscous drag force. For an init ial velocity v0 (at x = t = 0) and
a retarding force F = - bx nd t he velocity as a function of distance, v(x) , and show t hat the mass moves a fi nite distance before coming to rest.

Answer 1

The velocity v(x) as function of x is:

$\boxed{v(x)^2=v_0^2-\frac{bx^2}{m}}$

The finite distance moved by the mass is:

$\boxed{x=v_0\sqrt{\frac{m}{b}}}$

Explanation:

Given

Retarding force

$F=-bx$

Initial velocity

$u=v_0$

The mass is m

Acceleration due to retarding force will be

$a=\frac{F}{m}$

$\implies a=\frac{-bx}{m}$

We know that

$a=\frac{dv}{dt}$

or $a=\frac{dv}{dx}.\frac{dx}{dt}$

or, $a=v\frac{dv}{dx}$

Thus,

$v\frac{dv}{dx}=-\frac{bx}{m}$

$\implies mvdv=-bxdx$

$\implies m\int_{v_0}^{v(x)} vdv=-b\int_0^x xdx$

$\implies m\frac{v^2}{2}\Bigr|_{v_0}^{v(x)}=-b\frac{x^2}{2}\Bigr|_0^x$

$\implies m(\frac{v(x)^2}{2}-\frac{v_0^2}{2}=-b\frac{x^2}{2}$

$\implies \boxed{v(x)^2=v_0^2-\frac{bx^2}{m}}$

This is the expression of velocity v(x) as a function of distance x

Now if $v(x)=0$

Then

$0=v_0^2-\frac{bx^2}{m}$

$\implies x^2=\frac{mv_0^2}{b}$

$\implies x=\sqrt{\frac{mv_0^2}{b}}$

$\implies \boxed{x=v_0\sqrt{\frac{m}{b}}}$

Which is the finite distance moved by mass before coming to rest.

Hope this answer is helpful.

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