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# A child weighing 25 kg slides down a rope hanging from a branch of a tall tree. If the force of friction acting against him is 200N, the acceleration of child is ( $g = 10m/{s^2}$ ):(A) $22.5m/{s^2}$(B) $8m/{s^2}$(C) $5m/{s^2}$(D) $2m/{s^2}$

Last updated date: 20th Jun 2024
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Hint: As we know that the direction of frictional force is opposite of the direction of motion. Now make a FBD(free body diagram) of the given situation in which the child is sliding down the rope. And then calculate the net force (add forces like vector) acting on the child and then divide it with the mass of the child you will get the answer.

As we all know the basic formula of force:

$F = ma$
Where:
$F$ is the force on body
$a$ is the acceleration of body
$m$ is the mass of the body
Now see the free body diagram and observe the direction of force.
As force due to its weight that is $mg$ which is in downward direction and
Frictional force is in upward direction.
${F_{net}} = {F_{weight}} - {F_{friction}}$
${F_{weight}} = mg = 25 \times 10 = 250N$
${F_{net}} = 250 - 200$
${F_{net}} = 50$
$a = \dfrac{F}{m}$
$a = \dfrac{{50}}{{25}} = 2m/{s^2}$

Hence option D is correct.

Note: For such whenever there are multiple forces acting in different directions, for ease of calculation of the question make a free body diagram of the given situation and then calculate what is asked. While making a FBD of the figure do take directions of friction correctly. When the body is moving, friction acts in the opposite direction.