In figure, the blocks are at rest and a force of 10 N acts on the block of 4kg mass. The coefficient of static friction and the coefficient of kinetic friction are μs=0.2 and μk=0.15 for both the surfaces in contact. The magnitude of friction force acting between the surface of contact between the 2 kg and 4 kg block in this situation is
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
a
3N
b
4N
c
3.33 N
d
Zero
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
The minimum value of F required to be applied on the blocks to move is =μsm1+m2g=0.2×(2+4)×10=12N. Since the applied force is less than the minimum value of force required to move the blocks together, the blocks will remain stationary.