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Standard XII

Physics

Newton's Second Law

A fireman of ...

Question

A fireman of mass $60 k g$ slides down a pole. He is pressing the pole with a force of $600 N$ . The coefficient of friction between the hands and the pole is $0.5$ , with what acceleration will the fireman slide down $(g = 10 m / s^{2})$

1 m / s^{2}

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2.5 m / s^{2}

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10 m / s^{2}

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5 m / s^{2}

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Solution

The correct option is D $5 m / s^{2}$

Net downward acceleration $= \frac{Weight - Friction force}{Mass}$
$= \frac{m g - μ N}{m}$ where $N = 600 N$
$= \frac{60 \times 10 - 0.5 \times 600}{60}$
$= \frac{300}{60} = 5 m / s^{2}$

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A fireman of mass 60 kg slides down a pole. He is pressing the pole with a force of 600 N. The coefficient of friction between the hands and the pole is 0.5, with what acceleration will the fireman slide down (g=10 m/s2)

The correct option is D 5 m/s2 Net downward acceleration =Weight - Friction forceMass =mg−μNm where N=600 N =60×10−0.5×60060 =30060=5 m/s2

A fireman of mass $60 k g$ slides down a pole. He is pressing the pole with a force of $600 N$ . The coefficient of friction between the hands and the pole is $0.5$ , with what acceleration will the fireman slide down $(g = 10 m / s^{2})$

The correct option is D $5 m / s^{2}$

Net downward acceleration $= \frac{Weight - Friction force}{Mass}$
$= \frac{m g - μ N}{m}$ where $N = 600 N$
$= \frac{60 \times 10 - 0.5 \times 600}{60}$
$= \frac{300}{60} = 5 m / s^{2}$