A mass m sliding horizontally is subject to a viscous drag force. For

1.	A mass m sliding horizontally is subject to a viscous drag force. For an init ial velocity v0 (at x = t = 0) anda retarding force F = - bx find t he velocity as a function of distance, v(x) , and show t hat the mass moves afinite distance before coming to rest
Answer» 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 workdone is: Click Here: brainly.in/question/16376510