Question
Physics Question on laws of motion
Block A of mass 2kg is placed over block B of mass 8kg. The combination is placed over a rough horizontal surface. Coefficient of friction between B and the floor is 0.5. Coefficient of friction between A and B is 0.4. A horizontal force of 10N is applied on block B. The force of friction between A and B is (g=10ms−2)
A
100 N
B
40 N
C
50 N
D
zero
Answer
zero
Explanation
Solution
Total mass of blocks A and B=2+8=10kg
Friction between surface and combination of A and B
F=μR
=0.5×10×10=50N
Here applied force on box B is 10N that is less than 50N. So the system will be in rest because of this there is no friction between blocks A and B.