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A man slides down on a telegraphic pole with an acceleration equal to one-fourth on acceleration due to gravity . The frictional force between man and pole is equal to in terms to man's weight "w" : (A) W/4 (B) W/2 (C) 3W/4 (D) W?
Most Upvoted Answer
A man slides down on a telegraphic pole with an acceleration equal to ...
F=mg−ma
 =m(g−g/4)
 =3​/4mg
=3/4W
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A man slides down on a telegraphic pole with an acceleration equal to one-fourth on acceleration due to gravity . The frictional force between man and pole is equal to in terms to man's weight "w" : (A) W/4 (B) W/2 (C) 3W/4 (D) W?
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A man slides down on a telegraphic pole with an acceleration equal to one-fourth on acceleration due to gravity . The frictional force between man and pole is equal to in terms to man's weight "w" : (A) W/4 (B) W/2 (C) 3W/4 (D) W? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A man slides down on a telegraphic pole with an acceleration equal to one-fourth on acceleration due to gravity . The frictional force between man and pole is equal to in terms to man's weight "w" : (A) W/4 (B) W/2 (C) 3W/4 (D) W? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A man slides down on a telegraphic pole with an acceleration equal to one-fourth on acceleration due to gravity . The frictional force between man and pole is equal to in terms to man's weight "w" : (A) W/4 (B) W/2 (C) 3W/4 (D) W?.
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