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Standard XII

Physics

Friction General Understanding

A boy of mass...

Question

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 D $500 N$

The coefficient of friction $= 0.8, g = 10 m / s$ , mass $= 40 k g$
The frictional force $= μ N$
Hence, $μ N = m g \Rightarrow N = \frac{m g}{μ}$
$= \frac{40 \times 10}{0.8} = 500 N$

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A boy of mass 40kg 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=10m/s2, the horizontal force that he is applying on the pole is

The correct option is D 500NThe coefficient of friction =0.8,g=10m/s, mass =40kgThe frictional force =μNHence, μN=mg⇒N=mgμ=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 D $500 N$

The coefficient of friction $= 0.8, g = 10 m / s$ , mass $= 40 k g$
The frictional force $= μ N$
Hence, $μ N = m g \Rightarrow N = \frac{m g}{μ}$
$= \frac{40 \times 10}{0.8} = 500 N$