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 ................ $(m / s^{2}) (g = 10 m s^{- 1})$

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Solution

Net Force $= (250 - 50) N$
$= 200 N$
Acceleration $= \frac{F_{N e t}}{M} = \frac{200}{25} = 8 m / s^{2}$

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A boy of weight 25Kg 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 ................(m/s2)(g=10ms−1)

Net Force=(250−50)N=200NAcceleration =FNetM=20025=8m/s2

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 ................ $(m / s^{2}) (g = 10 m s^{- 1})$

Net Force $= (250 - 50) N$
$= 200 N$
Acceleration $= \frac{F_{N e t}}{M} = \frac{200}{25} = 8 m / s^{2}$