Questions
A 4 kg block A is placed on the top of 8 kg block B which rests on a smooth table.
A just slips on B when a force of 12 N is applied on A. Then, the maximum horizontal force F applied on
B to make both A and B move together, is
detailed solution
Correct option is B
When force is applied on A, acceleration produced will beFAmA+mB=μmAgmB …(i)When force is applied on B, acceleration produced will beFBmA+mB=μmAgmA …(ii)Dividing these two equations, we getFBFA=mBmA∴ FB=mBmA·FA=84×12=24 NTalk to our academic expert!
Similar Questions
A 40 kg slab rests on a frictionless floor as shown in the figure. A l0 kg block rests on the top of the slab. The static coefficient of friction between the block and slab is 0.60 while the kinetic friction is 0.40. The l0 kg block is acted upon by a horizontal force 100 N. If g = 9.8 , the resulting acceleration of the slab will be
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests