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TS EAMCET · Physics · Laws of Motion

A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and pole is equal to in terms of man's weight \(w\)

  1. A \(\frac{w}{4}\)
  2. B \(\frac{w}{2}\)
  3. C \(\frac{3 w}{4}\)
  4. D \(w\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{3 w}{4}\)

Step-by-step Solution

Detailed explanation

Man is sliding down the telegraphic pole with acceleration \(g / 4\). So, \(\begin{array}{rlrl} & m g-F & =\frac{m g}{4} \\ \Rightarrow & & F & =m g-\frac{m g}{4} \\ \Rightarrow & F & =\frac{3 m g}{4} \\ \Rightarrow & & F & =\frac{3 w}{4}\end{array}\)
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