JEE Main Physics question
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, what is the acceleration of the child (Take g = 9.8 m/s²)
Answer: (D)
Solution
Net downward force = mg − f = 25×9.8 − 2 = 245 − 2 = 243 N. Acceleration = 243/25 = 9.72 m/s².