A block of mass m is stationary with respect to a rough wedge is shown in figure. Starting from rest in
time $t\left(m=1 \mathrm{kg},\theta =30°,a=2{ \mathrm{ms}}^{-2},t=4 \mathrm{s}\right)$

Match the following columns for work done on the block and mark the correct option from the codes given below.

Column I (A)  By gravity (B)  By normal reaction (C) By friction (D) By all the forces

Column II (p)   144 J (q) 32 J (r )  $-$160J (s) 48 J

In  $t=4 \mathrm{s},v=at=8{ \mathrm{ms}}^{-1}\text{ and }s=\frac{1}{2}a{t}^2=16 \mathrm{m}$

$\mathrm{KE}=\frac{1}{2}m{v}^2=32 \mathrm{J}$

From work-energy theorem,
Work done by all the forces  $=\Delta KE=32J$

Work done by gravity,  $W_g=-mgh=-\left(1\right)\left(10\right)\left(16\right)=-160 \mathrm{J}$

Writing equation of motion, we have

$N\cos 30°+f\sin 30°-10=ma=2$

or    $\sqrt{3}N+f=24$     …(i)

$\Sigma F_x=0$

$\therefore N\sin 30°=f\cos 30°\text{ or }N=\sqrt{3}f$                 …(ii)

Solving Eqs. (i) and (ii), we have

$f=6 \mathrm{N} \text{ and } N=6\sqrt{3} \mathrm{N}$

Now, work done by normal reaction,  $W_N=\left(N\cos \theta \right)\left(s\right)$

$=\left(6\sqrt{3}\right)\left(\frac{\sqrt{3}}{2}\right)\left(16\right)=144 \mathrm{J}$

Work done by friction,  $W_f=\left(f\sin \theta \right)\left(s\right)=\left(6\right)\frac{1}{2}\left(16\right)=48 \mathrm{J}$

Work done by all the forces,

$W=W_g+W_N+W_F=-160+144+48=32 \mathrm{J}$

Hence,  $\mathrm{A}\to \mathrm{r},\mathrm{B}\to \mathrm{p},\mathrm{C}\to \mathrm{s},\mathrm{D}\to \mathrm{q}$