A sphere of mass m is set in motion with init | vedclass

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(A) The frictional force is given by $F = k x^n$.According to Newton's second law,the retardation is $a = -\frac{F}{m} = -\frac{k x^n}{m}$.Using the r

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$[\frac{m v_0^2 (n+1)}{2k}]^{1/(n+1)}$

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(A) The frictional force is given by $F = k x^n$.According to Newton's second law,the retardation is $a = -\frac{F}{m} = -\frac{k x^n}{m}$.Using the relation $a = v \frac{dv}{dx}$,we have $v \frac{dv}{dx} = -\frac{k x^n}{m}$.Integrating both sides with respect to their variables:$\int_{v_0}^{0} v \, dv = -\int_{0}^{x} \frac{k}{m} x^n \, dx$.Evaluating the integrals:$[-\frac{v^2}{2}]_{v_0}^{0} = -\frac{k}{m} [\frac{x^{n+1}}{n+1}]_{0}^{x}$.$0 - \frac{v_0^2}{2} = -\frac{k}{m} \frac{x^{n+1}}{n+1}$.$\frac{v_0^2}{2} = \frac{k x^{n+1}}{m(n+1)}$.Solving for $x$:$x^{n+1} = \frac{m v_0^2 (n+1)}{2k}$.$x = [\frac{m v_0^2 (n+1)}{2k}]^{1/(n+1)}$.

$A$ sphere of mass $m$ is set in motion with initial velocity $v_0$ on a surface where $kx^n$ is the frictional force,with $k$ and $n$ as constants and $x$ as the distance from the starting point. Find the distance at which the sphere will stop.

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