Question

A particle of mass $0.1kg$ is subjected to a force which varies with displacement as shown in fig.If starts its journey from rest at $x=0$,its velocity at $x=12m$ is:

Answer 1

Work done = area under $F-x$ curve

$W=10\times 4=40J$

$W=\Delta K.E$ (work energy theorem )

$\frac{1}{2}m{v}^2=W$

$V=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2\times 40}{0.1}}$

$v=20\sqrt{2}m/s$

Since, no force is acting after 4 sec, hence velocity will remain constant after 4 sec. Therefore, at $x=12$ m, it will have speed $v=20\sqrt{2}m/s$.