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Question

Physics Question on laws of motion

Block AA of mass 2kg2 \,kg is placed over block BB of mass 8kg8 \,kg. The combination is placed over a rough horizontal surface. Coefficient of friction between BB and the floor is 0.50.5. Coefficient of friction between AA and BB is 0.40.4. A horizontal force of 10N10\, N is applied on block BB. The force of friction between AA and BB is (g=10ms2)(g = 10\, m\,s^{-2})

A

100 N

B

40 N

C

50 N

D

zero

Answer

zero

Explanation

Solution

Total mass of blocks AA and B=2+8=10kgB=2+8=10\, kg
Friction between surface and combination of AA and BB
F=μRF =\mu R
=0.5×10×10=50N=0.5 \times 10 \times 10=50\, N
Here applied force on box BB is 10N10\, N that is less than 50N50\, N. So the system will be in rest because of this there is no friction between blocks AA and BB.