Question

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 200 N, the
acceleration of child is (g = 10 m/s2)
(1) 22.5 m/s2
(2) 8 m/s2
(4) 2 m/s2
(3) 5 m/s2
ka
horizontal​

Answer 1

Given:-

☆Mass of the child(m)=25kg

☆Frictional force acting against him=200N

☆Acceleration due to gravity(g)=10m/s²

To find:-

☆Acceleration of the child(a)

Solution:-

=>Weight of the child downwards:-

=>mg

=>25×10

=>250N

Now,net force(F)=(250-200)N=50N

By Newton's 2nd law of motion,we know that:-

=>F=ma

=>a=F/m

=>a=50/25

=>a=2m/s²

Hence,acceleration of the child is 2m/s² and correct option is (4).