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4-2.Friction
medium
A child weighing $25$ kg slides down a rope hanging from the branch of a tall tree. If the force of friction acting against him is $2\, N$, ........ $m/s^2$ is the acceleration of the child (Take $g = 9.8\,m/{s^2})$
A$22.5$
B$8$
C$5$
D$9.72$
Solution
(d) Net downward force = Weight -Friction
$\therefore $ $ma = 25 \times 9.8 – 2$ ==> $a = \frac{{25 \times 9.8 – 2}}{{25}} = 9.72\;m/{s^2}$
$\therefore $ $ma = 25 \times 9.8 – 2$ ==> $a = \frac{{25 \times 9.8 – 2}}{{25}} = 9.72\;m/{s^2}$
Standard 11
Physics
