Question

Force of 4N is applied on a body of mass 20 kg. The work done in $3^{rd}$ second is

Answer 1

$a=\frac{F}{m}=\frac{4}{20}=\frac{1}{5}m/s^2$
Distance moved in $3^{rd}sec=S_3-S_2$ $=\frac{1}{2}\times a\times 3^2-\frac{1}{2}\times a\times 2^2$ $=\frac{5}{2}a=\frac{5}{2}\times \frac{1}{5}=\frac{1}{2}m$
$\Delta W=f.s=4\times \frac{1}{2}=2J$