Question

A particle of mass ${10}^{-2}\text{kg}$ is moving along the positive x-axis under the influence of a force $F\left(x\right)=-\frac{K}{\left(2x^2\right)}$ where $K={10}^{-2}{\text{Nm}}^2$. At time $t=0$ it is at $x=1.0\text{m}$ and its velocity is $v=0$. Find magnitude of its velocity, when it reaches $=0.50\text{m}$. ( in $\text{m/sec})$

• A. 1.0
• B. 1
• C. 1.00

Answer 1

Answer: (A), (B), (C)

Given:
$m={10}^{-2}\text{kg}$
$k={10}^{-2}{\text{Nm}}^2$
$F=\frac{-k}{2x^2}$
$\Rightarrow F=-\frac{{10}^{-2}}{2x^2}$
$a=\frac{F}{m}=\frac{-{10}^{-2}}{\left({10}^{-2}\right)2x^2}$
$\Rightarrow a=\frac{-1}{2x^2}=v\frac{dv}{dx}$
$-{\int }_1^{0.5}\frac{dx}{2x^2}={\int }_0^{v}vdv$
$\Rightarrow -\frac{1}{2}{\int }_1^{0.5}{x}^{-2}dx=\frac{v^2}{2}$
$\Rightarrow \frac{-1}{2}{\left[\frac{-1}{x}\right]}_1^{0.5}=\frac{v^2}{2}$
$\Rightarrow \frac{1}{2}[\frac{1}{0.5}-1]=\frac{v^2}{2}$
$\Rightarrow |v|=1\text{m/s}$