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

Question

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 (A) (w/4) (B) (w/2) (C) (3 w/4) (D) w

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c02a79110343cd3b3629c902fe7d581d-.png" /> Man is sliding down the telegraphic pole with acceleration g / 4. So, m g-F=(m g/4) ⇒ F=m g-(m g/4) ⇒ F=(3 m g/4) ⇒ F=(3 w/4)

Q. 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 $w$

$\frac{w}{4}$

$\frac{w}{2}$

$\frac{3 w}{4}$

$w$

Man is sliding down the telegraphic pole with acceleration $g /4$ . So,
$m g - F = \frac{m g}{4}$
$\Rightarrow F = m g - \frac{m g}{4}$
$\Rightarrow F = \frac{3 m g}{4}$
$\Rightarrow F = \frac{3 w}{4}$

Q. 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

4w​

2w​

43w​

w

Man is sliding down the telegraphic pole with acceleration g/4. So, mg−F=4mg​ ⇒F=mg−4mg​ ⇒F=43mg​ ⇒F=43w​

Q. 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 $w$

$\frac{w}{4}$

$\frac{w}{2}$

$\frac{3 w}{4}$

$w$

Man is sliding down the telegraphic pole with acceleration $g /4$ . So,
$m g - F = \frac{m g}{4}$
$\Rightarrow F = m g - \frac{m g}{4}$
$\Rightarrow F = \frac{3 m g}{4}$
$\Rightarrow F = \frac{3 w}{4}$