Question

the displacement x of a particle moving in one dimension under the action of a constant force is related to time by the equation t=√ x + 3 where x in metre and t is in seconds the work done by the force in the first 6 seconds is​

Answer 1

The equation of motion of the particle is,

$\sf{\longrightarrow t=\sqrt x+3}$

$\sf{\longrightarrow \sqrt x=t-3}$

$\sf{\longrightarrow x=(t-3)^2\quad\quad\dots(1)}$

This is the displacement of the particle.

Work done by the force acting on it is,

$\displaystyle\sf{\longrightarrow W=\int F\ dx}$

Since the force is constant,

$\displaystyle\sf{\longrightarrow W=F\int dx}$

$\displaystyle\sf{\longrightarrow W=F\Big[x\Big]_{x_1}^{x_2}}$

[Here initial position and position after first 6 seconds are just taken as $\sf{x_1}$ and $\sf{x_2}$ respectively.]

From (1),

$\displaystyle\sf{\longrightarrow W=F\Big[(t-3)^2\Big]_0^6}$

$\displaystyle\sf{\longrightarrow W=F\Big[(6-3)^2-(0-3)^2\Big]}$

$\displaystyle\sf{\longrightarrow W=F\Big[3^2-(-3)^2\Big]}$

$\displaystyle\sf{\longrightarrow W=F\Big[9-9\Big]}$

$\displaystyle\sf{\longrightarrow\underline{\underline{W=0\ J}}}$