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

detailed solution

Correct option is D

In  t=4 s,v=at=8 ms-1 and s=12at2=16 mKE=12mv2=32 JFrom work-energy theorem,Work done by all the forces  =ΔKE=32JWork done by gravity,  Wg=-mgh=-(1)(10)(16)=-160 JWriting equation of motion, we haveNcos30°+fsin30°-10=ma=2or    3N+f=24     …(i)ΣFx=0∴  Nsin30°=fcos30° or N=3f                 …(ii)Solving Eqs. (i) and (ii), we havef=6 N   and   N=63 NNow, work done by normal reaction,  WN=(Ncosθ)(s)=(63)32(16)=144 JWork done by friction,  Wf=(fsinθ)(s)=(6)12(16)=48 JWork done by all the forces,W=Wg+WN+WF=-160+144+48=32 JHence,  A→r,B→p,C→s,D→q