Question

He only force acting on a 2 kg body as it moves along the positive x axis has component fx=-6x n where x is in metre. the velocity of the body at x=3m is 8 m/s, the velocity of the body at x=4 m is

Answer 1

The force acting on the body has the component,

$\longrightarrow\sf{F=-6x}$

As the force is varying with displacement, the work done by this force is,

$\displaystyle\longrightarrow\sf{W=\int-6x\ dx}$

$\displaystyle\longrightarrow\sf{W=-6\cdot\dfrac{x^2}{2}+c}$

$\displaystyle\longrightarrow\sf{W=-3x^2+c}$

This work done is stored in the body as its kinetic energy. Therefore,

$\displaystyle\longrightarrow\sf{K=-3x^2+c}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{2}\,mv^2=-3x^2+c}$

Since the mass of the body is 2 kg $\sf{(m=2\ kg),}$

$\displaystyle\longrightarrow\sf{v^2=-3x^2+c\quad\quad\dots(1)}$

Given that the body has velocity $\sf{8\ m\,s^{-1}}$ at $\sf{x=3\ m.}$ Then,

$\displaystyle\longrightarrow\sf{v^2=8^2}$

From (1),

$\displaystyle\longrightarrow\sf{-3(3^2)+c=64}$

$\displaystyle\longrightarrow\sf{-27+c=64}$

$\displaystyle\longrightarrow\sf{c=91}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf{v^2=-3x^2+91}$

Hence at $\sf{x=4\ m,}$

$\displaystyle\longrightarrow\sf{v^2=-3(4^2)+91}$

$\displaystyle\longrightarrow\sf{v^2=-48+91}$

$\displaystyle\longrightarrow\sf{v^2=43}$

$\displaystyle\longrightarrow\underline{\underline{\sf{v=\sqrt{43}\ m\,s^{-1}}}}$

$\displaystyle\longrightarrow\underline{\underline{\sf{v^2=6.56\ m\,s^{-1}}}}$