Answer
2.5 J
Work Step by Step
Force and displacement are in the same direction. The angle between them is 0.
$W=Fd\cos\theta=Fd\cos0^{\circ}=Fd$
Now, displacement $d=ut+\frac{1}{2}at^{2}$.
But initial velocity u=0 and acceleration $a=\frac{F}{m}$.
Therefore $d=\frac{1}{2}\times\frac{F}{m}\times t^{2}$
$\implies W=Fd=F\times\frac{1}{2}\times\frac{F}{m}\times t^{2}$
$=\frac{F^{2}\times t^{2}}{2m}$
Given: $F=5.0\,N$, $t=2\,s$ and $m=15\,kg$
Work done in 2 seconds $=\frac{(5.0\,N)^{2}\times(2\,s)^{2}}{2\times15\,kg}=3.33\,J$
Work done in the $2^{nd}$ second= (work done in 2 seconds)- (work done in first second)=$3.33\,J-0.83\,J=2.5\,J$
(we found the value of of the work done in first second to be equal to 0.83 J in the previous section)
