A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is ______ N. (Round off to the Nearest integer)(Take g=10ms−2)
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
answer is 30.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
FBD for wood,Horizontal , T=fL=μNVertical, N=N1+mg⇒T=μN1+mg........(1)FBD for manVertical, N1=m1g−T .........(2) put (2) in (1) ⇒T=μm1g−T +mg ⇒T=μm1g+mg1+μ=0.5×901.5=30N