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 find t he velocity as a function of distance, v(x) , and show t hat the mass moves a finite distance before coming to rest.

Answer 1

Answer:

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

Explanation:

Given,

$F = -bx$

⇒$m\frac{dv}{dt} = -bx$

⇒$m\frac{dv}{dx} .\frac{dx}{dt} = -bx$

But we know that,

$\frac{dx}{dt} = v$

⇒$mv\frac{dv}{dx} = -bx$

⇒$m.vdv=-b.xdx$

Let the final velocity be $v$

integrating on both sides,

$m\int\limits^v_{v_{0}} {v} \, dv =-b \int\limits^x_0 {x} \, dx$

⇒$m.\frac{v^{2}-v_{0} ^{2} }{2} =-b\frac{x^2}{2}$

⇒$v^{2} -v_{0}^{2} = \frac{-b}{m}x^{2}$

⇒$v^{2} = v_{0}^2 - \frac{bx^{2} }{m}$

We see that the body stops when $v=0$

⇒$0 = v_{0}^2 - \frac{bx^{2} }{m}$

⇒$x = v_{0}\sqrt{\frac{b}{m} }$

There the body stops after covering $x = v_{0}\sqrt{\frac{b}{m} }$  which is a finite distance

Answer 2

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.

Know More:

Q: When a force act on a body of mass its position x varies with time t as x=at^4+bt+c where a,b,c are constants work done is :

Click Here: https://brainly.in/question/16376510