A man of mass 65 kg slides down a telegraphic pole with an acceleration equal to (3/4) of acceleration due to gravity. The frictional force between the man and pole is N. (Take g =10 m / s 2 )

Question

A man of mass 65 kg slides down a telegraphic pole with an acceleration equal to (3/4) of acceleration due to gravity. The frictional force between the man and pole is N. (Take g =10 m / s 2 ) (A)  (B)  (C)  (D)

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ce43f27c9f2c762c7314cbc33231abc7-.png" /> Man is sliding down the telegraphic pole with acceleration (3 g /4). ∴ mg - F =(3 mg /4) ∴ F = mg -(3 mg /4) ∴ F =( mg /4) ∴ F =(65 × 10/4) ∴ F =162.5 N

Q. A man of mass $65 k g$ slides down a telegraphic pole with an acceleration equal to $\frac{3}{4}$ of acceleration due to gravity. The frictional force between the man and pole is_________ $N$ . (Take $g = 10 m / s^{2}$ )

Man is sliding down the telegraphic pole with acceleration $\frac{3 g}{4}$ .
$∴ m g - F = \frac{3 m g}{4}$
$∴ F = m g - \frac{3 m g}{4}$
$∴ F = \frac{m g}{4}$
$∴ F = \frac{65 \times 10}{4}$
$∴ F = 162.5 N$

Q. A man of mass 65kg slides down a telegraphic pole with an acceleration equal to 43​ of acceleration due to gravity. The frictional force between the man and pole is_________ N. (Take g=10m/s2 )

Man is sliding down the telegraphic pole with acceleration 43g​. ∴mg−F=43mg​ ∴F=mg−43mg​ ∴F=4mg​ ∴F=465×10​ ∴F=162.5N

Q. A man of mass $65 k g$ slides down a telegraphic pole with an acceleration equal to $\frac{3}{4}$ of acceleration due to gravity. The frictional force between the man and pole is_________ $N$ . (Take $g = 10 m / s^{2}$ )

Man is sliding down the telegraphic pole with acceleration $\frac{3 g}{4}$ .
$∴ m g - F = \frac{3 m g}{4}$
$∴ F = m g - \frac{3 m g}{4}$
$∴ F = \frac{m g}{4}$
$∴ F = \frac{65 \times 10}{4}$
$∴ F = 162.5 N$