Question

A particle of mass $0.5\text{kg}$ travels in a straight line with velocity $v=a{x}^{3/2}$, where $a=5$. What is the work done in joules by the net force during its displacement from $x=0$ to $x=2\text{m}$ ?

Answer 1

$W=\underset{0}{\overset{2}{\int }}mvdv=m{\left[\frac{v^2}{2}\right]}_0^2=0.5{\left[\frac{(a{x}^{3/2}{)}^2}{2}\right]}_0^2$
$=0.5{\left[\frac{(5x^{3/2}{)}^2}{2}\right]}_0^2=0.5{\left[\frac{25x^3}{2}\right]}_0^2$
$=0.5[\frac{25}{2}\times 8]=50\text{J}$

Why this question? Tip: Manupulations in the formula are done in a way so as to express it in variables, we have information about- $W=F.dx=ma.dx=mvdv$, we write $a$ as $vdv/dx$, since we have $v$ as a function of $x$)