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Second Law of Motion

A child weigh...

Question

A child weighing $25 k g$ slides down a rope hanging from a branch of a tall free. If the force of friction acting against him is $200 N$ , the acceleration of child is $(g = 10 m / s^{2})$

22.5 m / s^{2}

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8 m / s^{2}

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5 m / s^{2}

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2 m / s^{2}

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Solution

The correct option is D $2 m / s^{2}$
As child is sliding $d o w n$ so friction will act $u p w a r d$ that is opposite to $w e i g h t$ $m g = 25 \times 10 = 250 N$
So net downward force is $weight - friction = 250 N - 200 N = 50 N$ so acceleration $a = \frac{50 N}{25 K g} = 2 m s^{2}$ Option D is correct.

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