A boy of mass $40 k g$ is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is $0.8$ and $g = 10 m / s^{2}$ , the horizontal force that he is applying on the pole is

300 N

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400 N

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500 N

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600 N

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Solution

The correct option is C $500 N$
Here, horizontal force is $F$ and $f -$ frictional force

If boy moving constant speed, the

$f = m g$

$μ F = m g$

$0.8 F = 40 \times 10$

$F = \frac{40 \times 10}{0.8} = 500 N$

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A boy of mass 40 kg is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is 0.8 and g=10 m/s2, the horizontal force that he is applying on the pole is

The correct option is C 500 NHere, horizontal force is F and f− frictional forceIf boy moving constant speed, thef=mgμF=mg0.8F=40×10F=40×100.8=500N

A boy of mass $40 k g$ is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is $0.8$ and $g = 10 m / s^{2}$ , the horizontal force that he is applying on the pole is

The correct option is C $500 N$
Here, horizontal force is $F$ and $f -$ frictional force

If boy moving constant speed, the

$f = m g$

$μ F = m g$

$0.8 F = 40 \times 10$

$F = \frac{40 \times 10}{0.8} = 500 N$