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A man of mass 10kg is standing on a plank so that he is 20m from a point 'A'(point 'A' is not on plank).He walks 8m on the plank towards 'A' and stops.The mass of the plank is 40kg and frictipn between plank and ground is negligible.How far is the man from point 'A' now?
Dear Sindhuji P,
Ans:- Let the distance travelled by the man on the plank is = x mt, mass of the man= m kg and the plank=M kg
then the plank will move horizontally in the opposite direction a distance y.
So, the net displacement of the man is (x-y)
hence the eq is, m(x-y)=My
y=mx/(m+M)
given that x=8 m
hence y=1.6 mt. So the net displacement of the man is=(8-1.6)= 6.4 mt
Hence the present distance from point A is =(20-6.4)=13.6 mt (ans)
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All the best Sindhuji P !!!
Regards,
Askiitians ExpertsSoumyajit Das IIT Kharagpur
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