A boy of mass 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 (g=10ms–2)
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
10ms–2
b
12ms–2
c
8ms–2
d
2ms–2
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Net downward force acting on the boy = mg - fk ⇒ (25 x 10 - 50)N = 200N Acceleration =