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Standard XII

Physics

Newton's Second Law

A boy of mass...

Question

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 $m s^{- 2}$ , 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
For constant velocity
$F_{n e t} = 0$
Friction $= 40$ g
$μ N = 40$ g
$v = \frac{40 \times 10}{0.8}$
$= \frac{40 \times 10 \times 10}{8}$
$N = 500$ Newton.

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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=10ms−2, the horizontal force that he is applying on the pole is:

The correct option is C 500 NFor constant velocityFnet=0Friction=40gμN=40gv=40×100.8=40×10×108N=500 Newton.

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 $m s^{- 2}$ , the horizontal force that he is applying on the pole is:

The correct option is C 500 N
For constant velocity
$F_{n e t} = 0$
Friction $= 40$ g
$μ N = 40$ g
$v = \frac{40 \times 10}{0.8}$
$= \frac{40 \times 10 \times 10}{8}$
$N = 500$ Newton.