A block of mass m is stationary with respect to a rough wedge is shown in figure. Starting from rest intime $$tm=1$$ kg,θ$$=30$$°,$$a=2$$ $$ms-2$$,$$t=4$$ sMatch 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 forcesColumn $$II(p)$$ $$144$$ $$J(q)$$ $$32$$ $$J(r$$ $$)$$ $$-160J(s)$$ $$48$$ J

Question

A block of mass m is stationary with respect to a rough wedge is shown in figure. Starting from rest intime $$tm=1$$ kg,θ$$=30$$°,$$a=2$$ $$ms-2$$,$$t=4$$ sMatch 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 forcesColumn $$II(p)$$   $$144$$ $$J(q)$$ $$32$$ $$J(r$$ $$)$$  $$-160J(s)$$ $$48$$ J

Infinity Learn · Accepted Answer

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  $$=$$Δ$$KE$$$$=32JWork$$ done by gravity,  $$Wg=-$$$$mgh$$$$=-(1)(10)(16)=-160$$ JWriting equation of motion, we $$haveNcos30$$°$$+fsin30$$°$$-10=ma=2or$$    $$3N+f=24$$     …(i)Σ$$Fx=0$$∴  $$Nsin30$$°$$=fcos30$$° or $$N=3f$$                 …(ii)Solving$$ Eqs. $$(i) and (ii), we $$havef=6$$ N   and   $$N=63$$ NNow, work done by normal reaction,  $$WN=(Ncos$$θ$$)(s)=(63)32(16)=144$$ JWork done by friction,  $$Wf=(fsin$$θ$$)(s)=(6)12(16)=48$$ JWork done by all the forces,$$W=Wg+WN+WF=-160+144+48=32$$ JHence,  A→r,B→p,C→s,D→q

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