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Standard XII

Physics

1st Equation of Motion

A boy of weig...

Question

A boy of weight $25 k g$ slides down a rope hanging from the branch of a tree. If the force of friction against him is $50 N$ , the boy's acceleration is $(g = 10 m s^{- 2})$ .

10 m s^{- 2}

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12 m s^{- 2}

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8 m s^{- 2}

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2 m s^{- 2}

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Solution

The correct option is C $8 m s^{- 2}$
Simply balance the forces along with frictional force $.$
Therefore $,$
$F - f = m a$
$250 - 50 = 25 a$
$a = \frac{200}{25} = \frac{40}{5} = 8 m / s$
Hence,
option $(C)$ is correct answer.

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A boy of weight 25 kg slides down a rope hanging from the branch of a tree. If the force of friction against him is 50N, the boy's acceleration is (g=10 ms−2).

The correct option is C 8 ms−2Simply balance the forces along with frictional force.Therefore,F−f=ma250−50=25aa=20025=405=8m/sHence,option (C) is correct answer.

A boy of weight $25 k g$ slides down a rope hanging from the branch of a tree. If the force of friction against him is $50 N$ , the boy's acceleration is $(g = 10 m s^{- 2})$ .

The correct option is C $8 m s^{- 2}$
Simply balance the forces along with frictional force $.$
Therefore $,$
$F - f = m a$
$250 - 50 = 25 a$
$a = \frac{200}{25} = \frac{40}{5} = 8 m / s$
Hence,
option $(C)$ is correct answer.