Physics - NLM Question with Solution | TestHub

For P: When A and B move together, let their common acceleration be $a$. FBD for A: $f_{AB}=ma$. FBD for B: $F-f_{AB}=2ma$. Combining, $F=3ma$. Maximum static friction $f_{s,max}={\mu }_s{N}_A={\mu }_{s}mg=0.5mg=0.5m(10)=5m$. Thus, $F_{max}=3f_{s,max}=3(5m)=15m$. So (P)-(3). For Q and S: Since $F=18m>F_{max}=15m$, block A slips relative to block B. Kinetic friction $f_k={\mu }_k{N}_A={\mu }_{k}mg=0.4mg=0.4m(10)=4m$. For block A: $f_k=m{a}_A⟹4m=m{a}_A⟹a_A=4m/s^2$. So (S)-(4). For block B: $F-f_k=2m{a}_B⟹18m-4m=2m{a}_B⟹14m=2m{a}_B⟹a_B=7m/s^2$. So (Q)-(6). For R: The horizontal surface on which block B rests is smooth. Therefore, the friction force exerted by the ground on block B is zero. So (R)-(2).